Class 9 Maths - Lines And Angles

Draw a line segment $AB$ of $4$ cm in length. Draw a line perpendicular to $AB$ through $A$ and $B$ , respectively. Are these lines parallel?
If true write $1$ and if false then write $0$

The lines are perpendicular to a common line

$AB$ . Hence, the angle between these lines will be

$90^{\circ} + 90^{\circ} = 180^{\circ}$
Since, the angle between them is

$180^{\circ}$ , the lines are parallel to each other.

Can two acute angles be complement to each other always?
If true, then enter $1$ and if false, then enter $0$ .

We know, acute angle is the angle which is greater than

$\displaystyle { 0 }^{ o }$ and less than

$\displaystyle { 90 }^{ o }$ .

Two acute angles can be complementary to each other but it is not always true.

E.g.: Let us suppose two acute angles are

$70^o$ &

$30^o$ .
Here, the sum of these two acute angles will give us

$100^o$ .
Since, for being complementary, sum should be equal to

$90^o$ ,
we can say that not all pairs of acute angles are complementary.
Hence, the statement is false and

$0$ is the answer.

If true then enter $1$ and if false then enter $0$
Can two obtuse angles be adjacent angles?

If the sum of

$2$ angles is less than

$360^o$ , then they can be adjacent.

$\Rightarrow$ Since an obtuse angle is less than

$180^o$ , the sum of

$2$ obtuse angles will always be less than

$360^o$ .

Hence, two obtuse angles can be adjacent angles.

$\therefore$ The answer is

$1.$

If true then enter $1$ and if false then enter $0$
Can two right angles be supplementary?

Yes.

Two right angles are always supplementary to each other.
Since right angle=90 degrees, So the sum of two right angles will always be 180 degrees.

If true then enter $1$ and if false then enter $0$
Can two adjacent angles be supplementary?

Yes.

Two adjacent angles can be supplementary if their sum is 180 degrees. This is only possible if both angles are right angles or if one is acute and the other is an obtuse angle and their sum is 180 degrees.
E.g.

${ 110 }^{ o }+{ 70 }^{ o }={ 180 }^{ o }\\ { 90 }^{ o }+{ 90 }^{ o }={ 180 }^{ o }$

Measure of an angle is given below. Find the measure of it's complementary angle:
$58^o$ .

We know, two angles are complementary when they add up to

$90^o.$

Given, measure of one complementary angle is

$58^o.$

$\Rightarrow$ Measure of other complementary angle

$=90^o-58^o$

$=32^o.$

Hence, measure of the complementary angle of

$58^{o}$ is

$32^o$ .

If true then enter $1$ and if false then enter $0$
Can two acute angles form a linear pair?

No.

Acute angles can never form a linear pair.
For two angles to form a linear pair, they need to give a sum of 180 degrees.
But acute angles are less than 90 degrees, so two acute angles can never dd up to 180 degrees.

Measures of some angles are given below. Find the measures of their supplementary angles.

108 $^o$

Supplementary angles are two angles with a sum of

$180^o.$

One supplement angle is

$108^o.$

$\Rightarrow$ Other supplement angle

$= 180^o-108^o$

$=72^o.$

$\therefore$ Measure of a supplementary angle of

$108^{o}=72^o$

Measures of some angles are given below. Find the measure of their complementary angle:
$56$ $^o$

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $56^\circ$ .

Let the measure of its complement be $x^\circ.$

Then,

$\implies x + 56^o = 90^o$

$\implies x = 90^o - 56^o$

$\implies x = 34^\circ$ .

Hence, the complement of the given angle measures $34^\circ.$

Measures of some angles are given below. Find the measures of their complementary angles:
16 $^o$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $16^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-16^o$ $=74^o.$

$\therefore$ Measure of a complementary angle of $16^{o}=74^o$ .

Measures of some angles are given below. Find the measures of their complementary angles:
$68$ $^o$

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $68^{o}$ is,

$=90^{o}-68^{o}$

$=22^{o}$ .

Therefore, the complement angle of $68^{o}$ is $22^{o}$ .

Find the measure of the supplementary angle pair of 60 $^o$ . (in degrees)

Supplementary angles are two angles with a sum of

$180^o.$

Measure of one supplementary angle

$=60^o$

$\Rightarrow$ Measure of other supplementary angle

$= 180^o-60^o$

$=120^o.$

Measures of some angle is given below. Find the measures of its complementary angle:
If the angle is $\displaystyle \frac{2}{3}$ rd of a right angle.

We know, two angles are complementary when they add up to

$90^o.$

Given, measure of one complementary angle is $=\dfrac{2}{3}\times 90^o=60^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-60^o.$ $=30^o.$

$\therefore$ Measure of a complementary angle of $\dfrac{2}{3}\times 90^o=30^o.$ .

equal to its complement.(in $^\circ$ )

Let the measure of the required angle be

$x^{o}$ .

Since, the angle is equal to its own complement, both the angle and its complement will be equal.

Then,

$\implies x^{o}+x^{o}=90^{o}$

$\implies 2x=90^o$

$\implies x=\dfrac{90}{2}^o$

$\implies x=45^o$ .

Hence, the required angle measures

$45^{o}$ .

degrees $\\ 92^{\circ}, 27^{\circ},----$

Given two angles of the triangle,

$92^{\circ}, 27^{\circ}$

Let the third angle be

$x$
In a

$\triangle ABC$ , by angle sum property, sum of angles of a triangle is

$180^{\circ}$

$92^o + 27^o +x = 180^o$

$x = 61^o$
Third angle =

$61^{\circ}$

Side BC of a $\Delta ABC$ is produced in both the directions. Prove that the sum of the two exterior angles so formed is greater than $180^o$ .

If we extend

$BC$ on both ends, then the angles would be,

$\angle ACD=180-C$

$\angle ABF=180-B$

Sum of the

$2$ exterior angles

$=\angle ACD+\angle ABF$

$=(180-C)+(180-B)$

$=360-(B+C)$

$=360-(180-A)=180+A$

$\angle A$ is always positive.

$\therefore$ sum of the angles is greater than

$180°$ .

1003906_243958_ans_f2aa52e2982d4f9987e85299b0f176b1.png

When do you say two angles are adjacent?

Two angles are adjacent when they have a common side and a common vertex(corner point) and don't overlap

Observe the following figures and identify the type of angles in them

Angle between the arms is clearly less then

${ 90 }^{ \circ }$

So the angle

$C$ is acute angle.

Identify the following given angles as acute , right or obtuse.

Acute angle

$<90^0$ , Right angle

$=90^0$ ,Obtuse angle

$>90^0$

(i) acute , (ii) obtuse , (iii) Right , (iv) acute , (v) obtuse

In each of the following figure AB || CD. Find the value of x in each case.

Draw a line

$EF$ perpendicular to

$AB$ and

$CD$ passing through

$M$

$\triangle BME$

$\angle BME+\angle MBE+\angle BEM={180}^{\circ}$ (Angle sum proprerty)

$\angle BME+{ 35 }^{ \circ }+{ 90 }^{ \circ }={ 180 }^{ \circ }\\ \angle BME={ 180 }^{ \circ }-{ 125 }^{ \circ }\\ \angle BME={ 45 }^{ \circ }$

$\triangle DMF$

$\angle DMF+\angle MDF+\angle MFD={180}^{\circ}$ (Angle sum property )

$\angle DMF+{ 75 }^{ \circ }+{ 90 }^{ \circ }={ 180 }^{ \circ }\\ \angle DMF={ 180 }^{ \circ }-{ 165 }^{ \circ }\\ \angle DMF={ 15 }^{ \circ }$

$x=\angle DMF+\angle BME+\angle EMF$

$x={ 15 }^{ \circ }+{ 45 }^{ \circ }+{ 180 }^{ \circ }\\ x={ 240 }^{ \circ }$

Find the supplement of the following angles.
$58^{\circ}$

The supplement of an angle is the remaining angle left up to

$180^0$

So,

Supplement of

$58°=180-58=122°$

In the given figure if AB || DE, $\angle BAC = 35^o$ and $\angle CDE = 53^o$ , find $\angle DCE$ .

$\angle DCE=\angle EAB$ (Alternate angles as

$AB||DE$ and

$AE$ is transversal)

$\angle DEC={35}^{\circ}$

$\triangle DCE$ , we have

$\angle DCE+\angle CED+\angle CED={180}^{\circ}$ (Angle sum property)

$\angle DCE+35^{ \circ }+53^{ \circ }={ 180 }^{ \circ }\\ \angle DCE={ 180 }^{ \circ }-{ 88 }^{ \circ }=92^{ \circ }$

To prove: $\angle 1$ is to $\angle 8$ . Write reasons for the statement.

$\angle 5+\angle 8={180}^{\circ}$ -------(i) (Straight angles made on line

$p$ )

$\angle1=\angle5$ (Corresponding angles)

Put

$\angle 5=\angle1$ in (i)

$\Rightarrow \angle1+\angle8={180}^{\circ}$

Hence proved that

$\angle 1$ is supplementary to

$\angle8$

Statement Reason
$\angle 1 + \angle 8 = 180^o$ .......................

Reason

$:$

$\angle 1=\angle 5$ (Corresponding angles as

$l||m$ )

$\angle 5+ \angle 8={ 180 }^{ \circ }$ (Supplementary angles made on a line at a point)

Put

$\angle 5 = \angle 1$

$\angle 1+\angle 8 ={ 180 }^{ \circ }$

Identify the angles which are more than right angles and which are less than right angle?

Angles whose measure are less than

$90^o$ (right angle) are called as acute angles.

And the angles whose measure is more thean

$90^o$ (right angle) are called as obtuse angles.

1369166_694934_ans_01ea00cadd0e42a7a7a1e9d9616e1550.png

In each of the following figure AB || CD. Find the value of x in each case.

Draw a line

$GH$ parallel to

$AB$ and

$CD$ passing through

$E$

$\angle GEA=\angle EAB$ (Alternate angles)

$\angle GEA={65}^{\circ}$

$\angle GEC=\angle ECD$ (Alternate angles)

$\angle GEC={35}^{\circ}$

$\angle CEA=\angle GEA+\angle GEC$ (From figure)

$x={ 65 }^{ \circ }+35^{ \circ }={ 100 }^{ \circ }$

Find the complementary angle of $36^{\circ}$ .

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $36^\circ$ .

Let the measure of its complement be $x^\circ.$

Then,

$\implies x + 36^o = 90^o$

$\implies x = 90^o - 36^o$

$\implies x = 54^\circ$ .

Hence, the complement of the given angle measures $54^\circ.$

Can two adjacent angles be complementary ? Draw figure .

Yes.Because

$\angle AOB + \angle BOC=$

$90^0$

1076782_702911_ans_777266a208224888bbf2833abb53744b.png

Can two adjacent angles be supplementary ? Draw figure .

Yes. Because

$\angle AOC+ \angle COB=$

$180^0$

1076759_702908_ans_730536ecf957486ab1815d7082090a74.png

Write the measure of supplementary angle of the given angles:
$47^{\circ}$ .

$180^{\circ} - 47^{\circ} = 133^{\circ}$

Thus, Supplementary angle of 47^{\circ} is 133^{\circ}

Which of the following pairs of angles are supplementary ?

If sum the of two angles is

$180^0$ , then they are called supplementary angles.

So among the given figures,

$(i)$ and

$(ii)$ represents the pair of supplementary angles because in these cases the sum of angles are

$180^o$

Find the value of unknown ' $x$ ' in the following triangle:

Sum of angles of triangles =

$180^0$

$30^0+110^0+ x =180^0$

$x=40^0$

Fill up the table on finding complementary angle of given angle:
$Angle - 1$ $Angle - 2$
(1) $50^{\circ}$ _
(2) $63^{\circ}$
(3) $47^{\circ}$ _
(4) $56^{\circ}$ _
(5) $12^{\circ}$
(6) $67^{\circ}$ ___

We know that,

if two angles are complementary, then their sum is $90^o$ .

$(1)$ Given, the angle is $50^o$ .

Then, the complement of $50^{0}=90^{0}-50^{0}=40^{0}$ .

Hence, the complement of

$50^{0}$ is

$40^{0}$ .

$(2)$ Given, the angle is $63^o$ .

Then, the complement of $63^{0}=90^{0}-63^{0}=27^{0}$ .

Hence, the complement of

$63^{0}$ is

$27^{0}$ .

$(3)$ Given, the angle is $47^o$ .

Then, the complement of $47^{0}=90^{0}-47^{0}=43^{0}$ .

Hence, the complement of

$47^{0}$ is

$43^{0}$ .

$(4)$ Given, the angle is $56^o$ .

Then, the complement of $56^{0}=90^{0}-56^{0}=34^{0}$ .

Hence, the complement of

$56^{0}$ is

$34^{0}$ .

$(5)$ Given, the angle is $12^o$ .

Then, the complement of $12^{0}=90^{0}-12^{0}=78^{0}$ .

Hence, the complement of

$12^{0}$ is

$78^{0}$ .

$(6)$ Given, the angle is $67^o$ .

Then, the complement of $67^{0}=90^{0}-67^{0}=23^{0}$ .

Hence, the complement of

$67^{0}$ is

$23^{0}$ .

Write the measure of supplementary angle of the given angles:
$75^{\circ}$ .

$180^{\circ} - 75^{\circ} = 105^{\circ}$

Thus supplementary angle of given angle is

$105^{\circ}$

Measure of one angle of linear pair is given. Find the measure of another angle:
$111^{\circ}$ .

$180^{\circ} - 111^{\circ} = 69^{\circ}$

Thus, another angle is

$69^{\circ}$

Write the measure of the supplementary angle of the given angle:
$112^{\circ}$ .

Let the measure of the supplementary angle of

$112^\circ$ be

$x.$

As we know that the sum of supplementary angles is equal to

$180^\circ.$ So,

${112^ \circ } + x = {180^ \circ }$

$\Rightarrow x = {180^ \circ } - {112^ \circ }$

$\Rightarrow x = {68^ \circ }$

Thus, the supplementary angle of given angle is

$68^{\circ}.$

Write the measure of supplementary angle of the given angles:
$100^{\circ}$ .

Supplementry angle is sum of two angle give

$180^0$ .

One angles

$=100^0$ (given that)

other angle(supplement angle)=

$80^0$ .

Write the measure of supplementary angle of the given angles:
$81^{\circ}$ .

$180^{\circ} - 81^{\circ} = 99^{\circ}$

Thus, supplementary angle of given angle is

$99^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$109^{\circ}$ .

$180^{\circ} - 109^{\circ} = 71^{\circ}$

Thus supplementary angle of given angle is

$71^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$90^{\circ}$ .

$180^{\circ} - 90^{\circ} = 90^{\circ}$

Thus supplementary angle of given angle is

$90^{\circ}$

Write the measure of supplementary angle of the given angles:
$132^{\circ}$ .

$180^{\circ} - 132^{\circ} = 48^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$60^{\circ}$ .

$180^{\circ} - 60^{\circ} = 120^{\circ}$

Thus supplementary angle of given angle is

$120^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$20^{\circ}$ .

$180^{\circ} - 20^{\circ} = 160^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$130^{\circ}$ .

$180^{\circ} - 130^{\circ} = 50^{\circ}$

Thus, supplementary amngle of given angle is

$50^{\circ}$ .

Make a pair of angles forming linear pair from the following angles:
$27^{\circ}, 90^{\circ}, 130^{\circ}, 80^{\circ}, 35^{\circ}, 50^{\circ}, 145^{\circ}, 100^{\circ}, 90^{\circ}, 153^{\circ}$ .

Required Linear pairs angle are

$(90,90),(130,5),(27,153),(100,80)\& (35,145)$

$m\angle XVW = 120^{\circ}$ , then
(i) $m\angle WVY =$ and
(ii) $m\angle XVZ =$ .

(i)

$\angle$ WVY =

$180^{\circ} - 120^{\circ} = 60^{\circ}$ (

$\because$

$\angle$ WVY and

$\angle$ XVW are Supplementary angles. )
(ii)

$\angle$ XVZ =

$60^{\circ}$ (

$\because$

$\angle$ XVZ and

$\angle$ WVY are vertically opposite angles.)

How may pairs are formed of angles of linear pair?

Required pairs are formed as angles of Liner pair

$\angle PMX,\,\,\,and\,\,\,\angle XMQ$

$\angle PMY,\,\,\,and\,\,\,\angle YMQ$

Find the angles x and y in the figure. $AO=OB$

Since,

$AO=OB\Rightarrow \angle ABO=\angle x..........(1)$

Now,

$\angle AOB=92^o$ as it is vertically oppsite angle to

$92^o$

Also, According to Sum of Opposite angle Rule

$x+92^o=y...................(2)$

Also,

$\angle ABO+y=180^o\Rightarrow \angle ABO=180^o-y...................(3)$

From (1),(2) and (3)

$x=180^o-y\Rightarrow x=180^o-(x+92^o)\Rightarrow 2x=88^o\Rightarrow x=44^o$

$y=180^o-x=180^o-44^o=136^0$

Find x, y, z.

$x+120^\circ=180^\circ$ ----------angles on a straight line is equal to

$180^\circ$

$x=180^\circ-120^\circ=60^\circ$

$z=x=60^\circ$ -----------corresponding angles

$y=120^\circ$ -------------alternate angles

962073_1030560_ans_dec1876da84d42b7b312f8d4a74ec914.JPG

In the figures given, state whether the lines $l$ and $m$ are parallel or not.

(i)

$106^\circ + 67^\circ = 173^\circ$

Since sum of co-interior angles is not

$180^\circ$

Therefore given lines

$l$ and

$m$ are not parallel.

(ii) Alternate interior angle of

$75^\circ$

$=180^\circ - 75^\circ = 105^\circ$

Since alternate interior angles are not equal

Therefore, given lines

$l$ and

$m$ are not parallel.

(iii) Corresponding angle of

$57^\circ$

$=180^\circ - 123^\circ = 57^\circ$

Since, the corresponding angles are equal

Therefore the given lines

$l$ and

$m$ are parallel

Show that the angles of an equilateral triangle are $60^o$ each.

${\textbf{Step-1: Observe an equilateral triangle.}}$

${\text{We know that all the sides of a triangle are equal.}}$

$AB=BC=CA$

${\text{By angle sum property we know that the sum of all the angles of a triangle is}}$

$180^0.$

$\Rightarrow \angle A + \angle B + \angle C = 180^0$

${\textbf{Step-2: Find the angle of an equilateral triangle.}}$

${\text{All the angles of an equilateral triangle are equal, since all angles are opposite to equal sides.}}$

$\Rightarrow \angle A = \angle B = \angle C$

$\Rightarrow \angle A + \angle B + \angle C = 180^0$

$\Rightarrow \angle A + \angle A + \angle A = 180^0$

$\Rightarrow 3 \angle A = 180^0$

$\Rightarrow \angle A = 60^0$

$\Rightarrow \angle A = \angle B = \angle C=60^0$

${\textbf{Hence, the angles of an equilateral triangle are}}$ $60^0$

${\textbf{each.}}$

Can two obtuse angles be supplementary?

Two Angles are Supplementary when they add up to

$180^\circ$ .

Let us consider the least obtuse angle, i.e.,

$>90^\circ$ Say

$91^\circ$ and add it to itself.

$91^\circ+91^\circ=182^\circ$

This is

$>180^\circ$ . Hence, two obtuse angles can never be supplementary.

Find the supplement of $105.$

Supplement of

$105^\circ =180^\circ-105^\circ$

$=75^ \circ$

If $PQ\bot PS,\ PQ\parallel SR,\ \angle SOR = 28^{o}$ and $\angle QRT = 65^{o}$ , then find $x$ & $y$ .

$\because PQ\parallel SR$

$\angle PQR = \angle QRT$ (Alternate interior angle)

$\Rightarrow x + {28^ \circ } = {65^ \circ }$

$\Rightarrow x = {65^ \circ } - {28^ \circ }$

$\Rightarrow x = {37^ \circ }$

$In\,\,\,\Delta PQS$

$\angle P + \angle Q + \angle S = {108^ \circ }$ (By angle sum property)

$\Rightarrow {90^ \circ } + x + y = {180^ \circ }$

$\Rightarrow {90^ \circ } + {37^ \circ }+y = {180^ \circ }$

$\Rightarrow y = {108^ \circ } - 127$

$\Rightarrow y = {53^ \circ }$

Find the supplementary of $37^028'$ and $52^0 33'$ .

$For\quad an\quad angle\quad to\quad be\quad supplementary,the\quad sum\quad of\quad two\quad angles\quad should\quad be{ 90 }^{ \circ }.$

$Supplementary\quad of\quad { 37 }^{ \circ }{ 28 }^{ ' }=90-{ 37 }^{ \circ }{ 28 }^{ ' }={ 52 }^{ \circ }{ 32 }^{ ' }$

$Supplementary\quad of\quad { 52 }^{ \circ }{ 33 }^{ ' }=90-{ 52 }^{ \circ }{ 33 }^{ ' }={ 37 }^{ \circ }27^{ ' }$

In Fig. if $AB || DE, \angle BAC = 35^o$ and $angle CDE = 53^o$ , find $\angle DCE$ .

$\angle BAE=\angle CED$

$\therefore \theta =35^{\circ}$ (Here

$AB\parallel DE$ , Here AE acts as transverse)

$\Delta CDE\Rightarrow 53+35+\angle C=180$

$88+\angle C=180^{\circ}$

$\angle C=92^{\circ}$

$\therefore DCE=92^{\circ}$

1055494_1061992_ans_7a44f9a5d82845298d562cd4e1d4d9f7.PNG

If the sum of the measures of two angles is ${90^\circ}$ , then the angles are called _____ angles.

We know, when the sum of the measures of two angles is $90^\circ$ , the angles are called $\underline{\text{complementary}}$ angles.

Find the angles, when:
(i)The angles are complementary and the smaller is $40^ \circ$ less than the larger.
(ii)The angles are complementary and the larger is four times the smaller.

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

Then:

$i)$ If one angle is

$x$ , then its complement given is

$x+40^\circ$ .

Then,

$x+x+40^\circ=90^\circ$

$\implies$

$2x=90^o-40^o=50^o$

$\implies$

$x=\dfrac{90^o}{5}=18^o$ .

Then, smaller angle

$x=25^\circ$ and larger angle is

$x+40^o=25^o+40^o=65^\circ.$

$ii)$ If one angle is

$x$ , then its complement given is

$4x$ .

Then,

$x+4x=90^\circ$

$\implies$

$5x=90^o$

$\implies$

$x=\dfrac{90^o}{5}=18^o$ .

Then, smaller angle

$x=18^\circ$ and larger angle is

$4x=4\times18^o=72^o$ .

In the figure $\angle ABC={80}^{o}$ . Find $\angle DGC$ and $\angle DEF$ , given $AB||DE$ and $BC||EF$ .

Given

$\angle ABC=80^\circ$

$AB\parallel DE$ ,

$BC$ is transversal line,

Hence,

$\angle ABC=\angle DGC=80^\circ$

$BC\parallel EF$ ,

$ED$ is transversal line,

Hence,

$\angle DEF=\angle DGC=80^\circ$

Find $a.$

$a + {40^0} = {180^0}$ (linear pair)

$a = {180^0} - {40^0}$

$a = {140^0}$

1090268_1065558_ans_171fe7ccd24543c2bdf2d912dafc6240.png

Name the pairs of adjacent angles in the figures.

In the left fig. the pair of adjacent angles are

$1)\angle ANC,\angle CNB\\2)\angle CNB,\angle BNA\\3)\angle BNA,\angle ANC$

In the right fig. adjacent angles are

$\angle PQT,\angle PQR$

Manasa says, "Each angle in any pair of complementary angles is always acute". Do you agree? Give reason.

We know that, if two angles are complementary, then their sum is

$90^o$ .

$\therefore$ For a pair of angles to be complementary, both of them should be less than

$90°$ .

Also we know, angles less than

$90°$ are called acute angles.

Hence, Manasa is right, i.e., each angle in any pair of complementary angles is always acute.

$AE||BD, AC||ED$ & $AB=AC$ find $a,b,c$

1317500_1063589_ans_7ef60ade9b954bc0a34c410f162a4595.jpg

Find $\angle APB$ , if $AE || BR$ .

given

$AE\parallel BR$

Now AB acts as transversal.

$\angle A+\angle B=180^{\circ}$ (interior angles of same side of transversal are supplementary )

$\therefore 2x+2y=180^{\circ}$

$2(x+y)=180^{\circ}$

$x+y=90^{\circ}$

Now in

$\Delta APB$

$\angle PAB+\angle PBA+\angle APB=180^{\circ}$

$X+Y+\angle APB=180^{\circ}$

But

$x+y=90^{\circ}$

$\therefore \angle APB=180^{\circ}-(x+y)$

$\angle APB=180^{\circ}-90^{\circ}$

$\angle APB=90^{\circ}$

1062400_1062572_ans_d45ac64ff71a471db97db2731d5dea38.PNG

In the figure, $\Delta PQR$ is a right angled at Q, $\overline {ML} ||\overline {RQ}$ and $\angle LMR = {130^ \circ }$ . Find $\angle LPM$ and $\angle PML$

Given: $\angle Q = {90^0}$

$ML\parallel RQ$

$\angle LMR = {130^0}$

To find: $\angle LPM = ?$

$\angle PML = ?$

Statement

1. $\angle PML + \angle LMR = {180^0}$

$\Rightarrow \angle PML + {130^0} = {180^0}$

$\Rightarrow \angle PML = {50^0}$

2. $\angle PLM = \angle PQR$

$= {90^0}$

3. $\angle PLM + \angle PML + \angle LPM = {180^0}$

$\Rightarrow {90^0} + {50^0} + \angle LPM = {180^0}$

$\Rightarrow \angle LPM + {140^0} = {180^0}$

$\Rightarrow \angle LPM = {40^0}$

1114745_1074631_ans_b1974d33bdf844dfb75741b272ca513e.png

Find the measure of an angle which measures same as its complement.

Let $x$ be the measure of the angle.

Then, its $=90^o - x$ .

It is given that the measure of an angle that measures the same as its complement.

So, $x = 90^o - x$

$\implies$ $2x = 90^o$

$x = 45^{0}$ .

Hence, the measure of an angle that measures the same as its complement is $45^o$ .

AD is an altitude of an isosceles triangle ABC in which AB=AC show that, (i) AD bisects BC (ii) AD bisects $\angle A$ .

Given:-

$ABC$ is an isosceles

$\triangle$ .

$AB = AC$

$AD$ is altitude.

$\angle{ADB} = \angle{ADC} = 90°$

To prove:-

(i)

$AD$ bisects

$BC$ , i.e.,

$BD = CD$

(ii)

$AD$ bisects

$\angle{A}$ , i.e.,

$\angle{BAD} = \angle{CAD}$

Proof:-

$\triangle{ADB}$ and

$\triangle{ADC}$ ,

$\angle{ADB} = \angle{ADC} \; \left[ \text{Each } 90° \right]$

$AB = AC \; \left[ \text{Given} \right]$

$AD = AD \; \left[ \text{Common} \right]$

By R.H.S congruency,

$\triangle{ADB} \cong \triangle{ADC}$

By C.P.C.T.

$BD = CD$

$\angle{BAC} = \angle{CAD}$

Hence proved.

If $A,B,C$ are the angles of a triangle, show that
$\sin{B}\cos{(C+A)}+\cos{B}\sin{(C+A)}=0$

$A+B+C=\pi$

$\therefore \sin B \cos(C+A) +\cos B\sin(C+A)$

$=\sin(A+B+C)$

$=\sin\pi =0$

In $\triangle\ PQR,\angle\ Q=40^{o},\angle\ R=75^{o}$ , then find the shortest and the longest side of the triangle

$\longrightarrow \angle Q=40^o$

$\angle R=75^o$

$\therefore \angle P=180^o-(40^o+75^o)$

$\therefore \angle P=180-115$

$\therefore \angle P=65^o$

Side opposite to greatest angle is greatest and side opposite to shortest angle is shortest.

$\therefore$ side

$PR$ is shortest

$\therefore$ side

$PQ$ is longest

1374764_1161949_ans_0b211aa0856c40fcba0b14c1801041a4.png

In the adjoining figure, name the following pairs of angles.
Equal supplementary angles

From the figure,

$\angle BOE$ and

$\angle DOE$ are the equal supplementary angles because both equals to

$90^\circ.$

Hence, this is the answer.

Explain:
Can two acute angles be supplementary

No, If two angles are supplementry then their sum should be

$180^o$

but when two acute angles (largest acute angles) such that

$89^o+89^o=178^0<180^o$

hence, this is the answer.

Can two right angles be supplementary?

Yes because one right angle equals to

$90^\circ$ therefore two right angles sum up to

$180^\circ$ . Also, sum of supplementary angles sum up to

$180^\circ$ .

Hence, this is the answer.

Two complementary angles are in the ratio of $2:1$ . Find the measure of the angles.

It is given that two complementary angles are in the ratio $2:1$ .

Consider $2x$ and $x$ as the angles.

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, $2x + x = 90^0$

$\implies$ $3x = 90^0$

$\implies$ $x = \dfrac{90}{3} = 30$ .

Here, the angles will be $x=30^0$ and $2x=2 \times 30^o = 60^0$ .

In the adjoining figure, name the following pairs of angles.
Unequal supplementary angles

From the figure,

$\angle BOA$ and

$\angle DOA$ are unequal supplementary angles.

Similarly,

$\angle DOC$ and

$\angle COB$ are unequal supplementary angles.

Hence, this is the answer.

In the following figure, is $\angle 1$ adjacent to $\angle 2$ ? Given reasons.

No, $\angle 1$ and $\angle 2$ are not the adjacent angles.
Because,
$i)$ sum of $\angle1\ and\ \angle2$ doesn't equal to $180^{\circ}$
$ii)$ They have no common vertex.

1065822_1090336_ans_1b0b22d7e77840b299757d6e67f546e6.png

Find the angles of a triangle that are in ratio $2 : 2 : 5$ .

Given angles are in ratio

$2:2:5$

Let the angles be

$2x, 2x$ and

$5x$

sum of angles

$=180^{\circ}$

$\implies 2x+2x+5x=180^{\circ}$

$\implies 9x=180^{\circ}$

$\implies x=20^{\circ}$

$\therefore$ angles are

$(2x, 2x, 5x)$

$\implies (40^{\circ}, 40^{\circ}, 100^{\circ})$ .

Find the value of $x$

Let fourth angle be

$M$

$\angle M=180°-105°\\ \quad=75°$

$y+M=180°\\y+75°=180°\\ \Rightarrow y=180°-75°\\y=105°\\x=75°$

(Opposite angle of parallelogram)

$z=105°$

(Opposite angles of parallelogram)

Two adjacent angle of $||gm$ are $2x-15^o$ and $7x-75^o$ .Find all angles of $||gm$ ?

Adjacent angles of a parallelogram add up to

$180^o$ , using that:

$2x-15^o+7x-75^o=180^o\implies9x-90^o=180^o$

$\implies 9x=270^o\implies x=30^o$

Therefore, the angles are

$2x-15^o=2\cdot30^o-15^o=45^o$
Other ange =

$180^o-45^o=135^o$

As opposite angles of a parallelogram are equal, the other two angles are also

$45^o$ and

$135^o$ .

Find each one of lettered angles giving reason

$a=70^o$ [Alternate interior angles are equal]

1368943_1177624_ans_71a672231efe4119b23766f10262dec1.png

Find x in the following figure $(AB || CD)$ :

Since,

$AB \parallel CD$ , therefore,

$x+3x=180^{o}$

$[Interior\ angles]$

$4x=180^{o}$

$x=45^{o}$

Find the complements of the following angles:-
a) ${45^ \circ }$
b) ${8^ \circ }$
c) ${20^ \circ }$
d) ${25^ \circ }$
e) ${78^ \circ }$

We know that, if two angles are complementary, then their sum is

$90^o$ .

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

a) The complement of

$x={45 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 45 }^{ o }={ 45 }^{ o }.$

b) The complement of

$x={ 8 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 8 }^{ o }={ 82 }^{ o }.$

c) The complement of

$x={ 20}^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 20 }^{ o }={ 70 }^{ o }.$

d) The complement of

$x={25 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 25 }^{ o }={ 65 }^{ o }.$

e) The complement of

$x={ 78 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 78 }^{ o }={ 12 }^{ o }.$

Calculate the value of $x :$

(a) straight line

$\Rightarrow 2x+3x= 180^{\circ}$

$5x= 180^{\circ}$

$x= 36^{\circ}$

(b) Right angle

$\Rightarrow 5x+4x= 90^{\circ}$

$9x= 90^{\circ}$

$x= 10$

$(c) 2x+x+4x+3x= 360^{\circ}$

$10x= 360^{\circ}$

$x= 36$

$(d) 3x-5+x+20+65^{\circ}= 180^{\circ}$

$4x+80= 180^{\circ}$

$4x=100$

$x= 25$

$(e)2x+x+60+2x= 360$

$5x=300$

$x= 60^{\circ}$

Find the value of $x$ ?

Given that,

$\Delta$ ABC

$\angle ACD+\angle ACB=\angle BCD$

$120^o+\angle ACB=180^o$ (angle of lines)

$\angle ACB=180^o-120^o$

$\angle ACB=60^o$

Now, given that,

$AB=AC$

So,

$\angle ACB=\angle ABC$

$\angle ABC=60^o$

$x=60^o$

Hence, this is the answer.

1197400_1242786_ans_a489d6287e6a4c4b846f4ccddfd9edc6.png

Find x in the following figures :

Since,

$PQ \parallel RS$ , therefore,

$3x=90^{o}$

$[corresponding \ angles]$

$x=30^{o}$

In the figure $l\parallel m$ and $p$ is a transversal, find the value of $x$ , giving necessary reasons.

Given:

$l\parallel m$ and

$p$ is the transversal.

When two parallel lines are cut by a transversal, the resulting corresponding angles are congruent.

Here,

$5x$ and

$2x+24$ are the corresponding angles.

$\Rightarrow 5x=2x+24$

$\Rightarrow 3x=24$

$\Rightarrow x=8$

In figure given alongside $AB$ and $CD$ are intersecting lines. Name the following pair of angles.
Adjacent angles that do not form a linear pair and each less than $180^{o}$ .

$(\angle AOC$ ,

$\angle COE)$ ,

$(\angle COE$

$\angle EOB)$ ,

$(\angle BOE$ ,

$\angle BOD)$ are the required pairs.

From the adjoining figure find $x$ , i $AOB$ is a straight line. Hence complete the following
Which is an obtuse angle?

R.E.F image

Now we know

$\angle POA +\angle POB =180^{\circ}$ [linear pair]

$x+10+x-10=180^{\circ}$

2x=180

$[x=90^{\circ}]$

Now

$\angle POA$ =x+10=

$100^{\circ}$

$\angle POB=x-10=90-10=80$

$\therefore \angle POA$ obtuse angle.

1215404_1368967_ans_336e8beb05574e48a0fadc33b9c5dcc4.JPG

The supplementary angle of $40^{o}$ is

Supplementary of 40 degrees is

$180 - 40 = 140^\circ$

From the adjoining figure find $x$ , i $AOB$ is a straight line. Hence complete the following
Which is an acute angle?

R.E.F image

now ,we know

$\angle POA+\angle POB=180^{\circ}$

$x+10+x-10=180^{\circ}$

$2x= 180^{\circ}$

$[x=90^{\circ}]$

NOW

$\angle POB=90+10=80$

$\angle POA=90+10=100^{\circ}$

$\angle$ POB is acute angle.

1215405_1368972_ans_e287b73be2794de99d8d5f67094481f3.JPG

Define adjacent angle.

Adjacent angles are two angles that have a common vertex and a common side. The vertex of an angle is the endpoint of the rays that form the sides of the angle.

Fill in the blank.
The sum of the measure of supplementary angle is ______

complimentary angles are angles

whose sum is

$90^{\circ}$

where as

supplementary angles are angles whose sum is

$180^{\circ}$

What do you mean by linear pair of angles ?

Two adjacent angles form a linear pair of angle, if their non-common arms are two opposite rays. Sum of angles of linear pair is always

$180^{\circ}$ .

Find the supplement of the angle whose measure is ${60}^{\circ}$

The sum of angles whose measure is

${180}^{\circ}$ are supplementary angles.

Supplement of

${60}^{\circ}={180}^{\circ}-{60}^{\circ}$

$={120}^{\circ}$

In figure given alongside $AB$ and $CD$ are intersecting lines. Name the following pair of angles.
Equal supplementary angles

Equal supplementary angles:

$\angle(AOE)$ and

$\angle(BOE)$

In Fig. If $x+y=w+z$ then prove that $A O B$ is a line.

Given

$x+y=z+w\cdots(1)$

From Angle at a point theorem

Sum of all angles at a point is

$360^{\circ}$

$x+y+z+w=360\\x+y+x+y=360\\2(x+y)=360\\x+y=180$

$x,y$ are adjacent angles and form

$180^{\circ}$ together so

$AOB$ is a straight line

If the ratio between two complementary angles $2:3$ , then find the angles.

It is given that two complementary angles are in the ratio $2:3$ .

Consider $2x$ and $3 x$ as the angles.

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, $2x + 3x = 90^0$

$\implies$ $5x = 90^0$

$\implies$ $x = \dfrac{90}{5} = 18^0$ .

Here, the angles will be $2x= 2\times18^0=36^o$

and $3x=3\times 18^o = 54^0$ .

Find the supplementary angle of $105^{o}$ .

Two angles are said to be supplementary to each other if they sum up to

$180^o$ .

Now the supplementary angle of

$105^{o}$ is

$180^o-105^o=75^o$ .

In the figure.

$65+x+x=180^o$

(linear Angle)

$\Rightarrow 2x=180-65$

$\Rightarrow 2x=115$

$\Rightarrow x=57.5^o$ .

1205643_1449644_ans_bea9393ae86f4207a2eb21de3230f7ce.jpg

Find the supplementary angle of $115^{o}$ .

Two angles are said to be supplementary if they sum up to

$180^o$ .

Now the supplementary angle of

$115^{o}$ is

$180^o-115^o=65^o$ .

Find the supplementary angle of $87^{\circ}$

Let

$x$ be the supplementary angle of

$87°$

Two angles are called supplementary when their measures add up to $$180$$ degrees.

Therefore,

$\Rightarrow x + 87° = 180°$

$\Rightarrow x = 180° - 87°$

$= 93°$

Hence the supplementary angle of

$87°$ is

$93°$

Find the supplementary angle of $40^o$ .

Two angles are said to be supplementary if they sum upto

$180^o$ .

Now supplementary angle of

$40^o=180^o-40^o=140^o$ .

For an angle ${ x }^{ 2 }$ , find: The complementary angle.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $x^2$ .

Then, its complement is $90^o-x^2$ .

Find the measure of an angle which is $36^\circ$ more than its complement.

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

Given, an angle is $36^o$ more than its complement.

$\Rightarrow$ $x=(90^o -x)+36^o$

$\Rightarrow$ $x=90^o -x+36^o$

$\Rightarrow$ $x+x=126^o$

$\Rightarrow$ $2x=126^o$

$\Rightarrow \ x=\dfrac {126^o}{2}$

$\Rightarrow \ x=63^o$ .

Therefore, complementary of $x$ ,

$=90^o -x$

$=90^o -63^o$

$=27^o$ .

Hence, the required angle is $x=63^o$ .

Classify the following angle as right, straight, acute, obtuse or reflex.

Step 1: To classify

(1) Given angle as right, straight, acute, obtuse or reflex

Step 2: Classification

The given angle shown in the figure is Straight angle as its measure is

$180^o$ .

Classify the following angle as right, straight, acute, obtuse or reflex.

Step 1: To classify

(1) Given angle as right, straight, acute, obtuse or reflex

Step 2: Classification

If the angle is less than

$90^o$ then it is an acute angle and here we can see that the angle measures less than

$90^o$ so it's an acute angle.

Draw a rough sketch of a quadrilateral KLMN. State two pairs of adjacent angles.

Two angles of a quadrilateral having a common arm are called its adjacent angles.

Therefore,

$\angle K, \angle L$ and

$\angle M$ ,

$\angle N$ are the pair of adjacent angles.

$\angle K, \angle N$ and

$\angle L, \angle M$ are the pair of adjacent angles.

1337509_1644334_ans_627d70ab706f466d967c136a38f8af77.png

Write down the measures of some acute angles.(give at least two examples)

Step: 1

Explanation:

An angle is said to be acute angle if it is lesser than

$90^o$

$45^o, 70^o$ . are example of acute angle as it measure less than

$90^o$

Find five situations where obtuse angles are made.

Step 1: To find

(1) Five situations where obtuse angles are made

Step 2: Five situations

Obtuse angle examples are as follows:

The open book kept for reading, Wide open cupboard door, Rooftop, Clock showing 8 o’clock, the base of a Tripod.

In the given figure, $\angle ABC$ and $\angle PQR$ are given such that $AB\parallel PQ$ and $BC\parallel QR$ . Find the value of $\angle x$ and $\angle y$

1308540_1530401_ans_43ecb8c25837443694a11b4deb85c43e.jpeg

Write down the measures of some obtuse angles.(give at least two examples)

Step 1: To write

(1) Name of two obtuse angles

Step 2: Names

$105^o, 132^o$ are two examples of obtuse angle as they measure greater than

$90^o$ and less than

$180^o$ .

List ten situations where the angles made are acute.

Acute angle examples

Scissor, Divider, Pizza slice,Hands of clock,Cloth Hanger,Spokes of bicycle,Tip of pencil,Open notebook,Alphabets (A,M,N),

Identify whether the following pairs of angles are complementary or supplementary:
$45^o, 45^o$

If sum of two angles is

$180^o$ , then they are called supplementary angles. If sum of two angles is

$90^o$ . then they are called complementary angles.

$45^o + 45^o =90^o$ These are complementary angles.

Identify whether the following pairs of angles are complementary or supplementary:
$112^o,68^o$

If sum of two angles is

$180^o$ , then they are called supplementary angles. If sum of two angles is

$90^o$ . then they are called complementary angles.

$112^o + 68^o =180^o$ These are supplementary angles.

Fill in the blanks:
If two adjacent angles are supplementary, they form a ______.

If two adjacent angles are supplementary, they form a linear pair.

Identify whether the following pairs of angles are complementary or supplementary:
$63^o, 27^o$

If sum of two angles is

$180^o$ , then they are called supplementary angles. If sum of two angles is

$90^o$ . then they are called complementary angles.

$63^o + 27^o =90^o$ These are complementary angles.

Fill in the blanks:
If two angles are supplementary, then the sum of their measures is ______.

If two angles are supplementary, then the sum of their measures is

$\underline{\,\,180^o\,\,}$ .

Identify whether the following pairs of angles are complementary or supplementary:
$130^o, 50^o$

If sum of two angles is

$180^o$ , then they are called supplementary angles. If sum of two angles is

$90^o$ . then they are called complementary angles.

$130^o + 50^o =180^o$ These are supplementary angles.

Fill in the blanks:
If two lines intersect a point, then the vertically opposite angles are always ______.

If two lines intersect a point, then the vertically opposite angles are always equal.

Find the value of $x$ .

Solution:-

Since both lines are parallel.

Hence, by alternate angles property,

$x = 47$

Find the complement of the following angle: $35^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $35^o$ .

Then, the complement of $35^{0}=90^{0}-35^{0}=55^{0}$ .

In Fig., write all pairs of adjacent angles and all the linear pairs.

1653470_1669234_ans_3163c84ce76847afb1dd023ccb153ff8.jpg

Find the complementary angle of the following angle: $85^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $85^o$ .

Then, the complementary angle of $85^{0}=>90^{0}-85^{0}=5^{0}$ .

Fill in the blanks in each of the following to make the statement true:
Two lines parallel to the same line are ______ to each other.

1656341_1669386_ans_46d77c8980784e52934f2c888e19e6e9.jpg

In Fig., find $x$ . Further find $\angle BOC, \angle COD$ and $\angle AOD.$

1653475_1669243_ans_8c93588fa5d04c1d91acd5ffc3f06ff3.jpg

If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.

1656325_1669330_ans_6f89c09f91824aeba9c68c186d5e4c91.jpg

Find the supplement of the following angle.
$70^o$ .

We know that, pair of angles are said to be supplementary angles if their sum is equal to 180

$^\circ$ .

Let,

$x$ be the supplement angle for 70.

so,

$70^\circ+x=180^\circ$

$=> x=110^\circ$

Identify the following pairs of angles are complementary or supplementary:
$80^o,10^o$

If sum of two angles is

$180^o$ , then they are called supplementary angles. If sum of two angles is

$90^o$ . then they are called complementary angles.

$80^o +10^0 = 90^o$ These are complementary angles.

Find the supplement of the following angle:
$90^o$ .

We know that, pair of angles are said to be supplementary angles if their sum is equal to 180

$^\circ$ .

Let,

$x$ be the supplement angle for

$90^o$ .

So,

$90^\circ+x=180^\circ$

$\Rightarrow x=90^\circ$

Hence, supplement angle of

$90^\circ$ is

$90^\circ$ .

Can two angles be supplementary, if both of them be acute?

No, because sum of two acute angles are always less than

$180^o$

Identify the complementary and supplementary pairs of angles from the following pair.
$100^o, 80^o$ .

We know that:

Sum of pair of complementary angle is

$90^o$

Sum of pair of supplementary angle is

$180^o$

As,

$100^o+80^o = 180^o$

$\therefore$ Given pair of angles are supplementary angles.

Can two angles be supplementary, if both of them be obtuse?

No, because sum of two obtuse angles will be more than

$180^o$ .

Can two angles be supplementary, if both of them be right?

Yes, as sum of two right angles is

$180^o$ .

$63^o, 27^o$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $63^o$ and $276^o$ .

Then, their sum $=63^o+27^o=90^o$ .

Therefore, they are complementary.

Find the supplement of the following angle.
$135^o$ .

We know that, pair of angles are said to be supplementary angles if their sum is equal to 180

$^\circ$ .

Let,

$x$ be the supplement angle for

$135^o$ .

so,

$135^\circ+x=180^\circ$

$=> x=45^\circ$

Identify the complementary pair of angles from the following pair:
$25^o, 65^o$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $25^o$ and $65^o$ .

Then, their sum $=25^o+65^o=90^o$ .

Therefore, they are complementary.

Find the supplement of the following angle.
$120^o$ .

We know that, pair of angles are said to be supplementary angles if their sum is equal to 180

$^\circ$ .

Let,

$x$ be the supplement angle for 120.

so,

$120^\circ+x=180^\circ$

$=> x=60^\circ$

$215^{o}$

Here, $215^{o}$ is an angle which is more than $180^{o}$ but less than $360^{o}$ .

Hence, it is a reflex angle.

Classify the angles whose magnitudes are given below:
$89^{o}$

Acute angle

Classify the angles whose magnitudes are given below:
$180^{o}$

Straight angle

Draw rough diagrams of two angles such that they have
One point in common.

$\angle AOB$ and

$\angle BOC$ have one point

$O$ in common.

$328^{o}$

Here, $328^{o}$ is an angle which is more than $180^{o}$ but less than $360^{o}$ .

Hence, it is a reflex angle.

Classify the angles whose magnitudes are given below:
$125^{o}$

Obtuse angle

Fill in the blanks:
Two different lines in a plane either ...... at exactly one point or are parallel.

Two different lines in a plane either intersect at exactly one point or are parallel.

Classify the angles whose magnitudes are given below:
$56^{o}$

Acute angle

State which of the following angles with a small letter in the following diagrams are acute, which obtuse and which are reflex:

We know:

1. An angle which is less than

$90^{o}$ called an acute angle.

2. An angle which is more than $90^{o}$ but less than $180^{o}$ called an obtuse angle.

3. An angle which is more than $180^{o}$ but less than $360^{o}$ is called a reflex angle.

Here,

$\angle a$ and

$\angle c$ are acute,

$\angle h$ is obtuse,

3. reflex

$\angle a$ , reflex

$\angle b$ and

$\angle c$ are reflex angles.

Fill in the blanks
An angle whose measure is less than that of a right angle is ......

An angle whose measure is less than that of a right angle is acute.

Classify the angle whose magnitude is given below:
$178^{\circ}$

The given angle is

$178^{\circ}$

We know, the angles larger than

$90^{\circ}$ and smaller than

$180^{\circ}$ are called obtuse angles.

Hence, the angle

$178^{\circ}$ is an obtuse angle.

Classify the angle whose measure is given below:
$125^{\circ}$

$\because 90^{\circ}<125^{\circ}<180^{\circ}$

$\implies$

$125^{\circ}$ is an obtuse angle

Classify the angles whose measures are given below:
$56^{o}$

Acute angle

Name the following triangles with regards to angles:

The given triangle is an obtuse-angled triangle as

$\angle B$ is obtuse and greater than

$90^{o}$ .

Classify the angles whose measures are given below:
$180^{o}$

Straight angle

Write down each pair of adjacent angles shown in the above figures:

$\angle PQT$ and

$\angle PQR;\ \ \angle ORU$ and

$\angle QRP; \ and\ \angle RPS \ and \ \angle RPQ$ are adjacent angles.

If sum of measures of two angles is $180^{o}$ , then the are _______.

Supplementary angles are two angles whose sum is 180 degrees

The supplement of an acute is always ________ angle.

We know that two angles are supplementary when they add up to

$180^o$ .

If one of the angles is an acute angle i.e., less than

$90^o$ , its supplement must be more than

$90^o$ so that their sum becomes

$180^o$ .

Hence, the supplement of an acute is always

$\text{obtuse}$ angle.

Two lines in a plane which do not meet at a point anywhere are called ________ lines.

Two lines in a plane which do not meet at a point anywhere are called

$\text{parallel}$ lines.

Write down each pair of adjacent angles shown in the above figures:

$\angle TSV$ and

$\angle USV;\ \ \angle SVT$ and

$\angle SVU;$ are adjacent angles.

Two angles forming a ________ pair are supplementary.

In diagram,

$\angle POQ$ and

$\angle ROQ$ are forming a linear pair and they are supplementary.

Hence,

Two angles forming a

$\text{ linear }$ pair are supplementary.

1671831_1789953_ans_ff12420e227f4b61950339f413e890d1.png

If two angles have a common vertex and their arms form opposite rays Fig., Then, write all the pairs of vertically opposite angles.

Two pairs of vertically opposite angles are

$\angle 1$ and

$\angle 3, \angle 2$ and

$\angle 4$ .

In each of the above figure write, if any, $(i)$ each pair of vertically opposite angles, and $(ii)$ each linear pair.

Here, no pair of vertically opposite angles are there.

And pairs of linear pair are:

$\angle ABD$ and

$\angle DBC, \angle ABE$ and

$\angle EBC$

Write down each pair of adjacent angles shown in the above figures:

$\angle AOC$ and

$\angle AOD;\ \ \angle BOC$ and

$\angle BOD; \ \angle AOC\ and\ \angle BOC; \ \angle AOD \ and\ \angle BOD$ are adjacent angles.

The drawing below Fig., show angles formed by the goalposts at different positions of a football player.The greater the angle, the better chance the player has of scoring a goal. For example, the player has a better chance of scoring a goal from Position $A$ than from POsition $B$ .
In Parts (a) and (b) given below it may help to trace the diagram and draw and measure angles.
Estimate atleast two situations such that the angles formed by different positions of two player are complement to each other.

Two player are in such types of different positions that they formed two pairs of angle'

$(45^o, 45^o)$ and

$(60^o, 30^o)$ ' that are a complement to each other

$(\therefore 45^o + 45^o=90^o)$ and

$(60^o, 30^o=90^o)$

Name the pairs of supplementary angles in the above figure:

Pairs of supplementary angles are

$\angle POS$ and

$\angle SOQ, \angle SOQ$ and

$\angle QOR, \angle QOR$ and

$\angle ROP, \angle ROP$ and

$\angle POS$

If two angles have a common vertex and their arms form opposite rays Fig., Then, how many angles are formed?

Total

$13$ angles will be formed.

In Fig., $OR \bot \ OP$ .
Name all the pairs of adjacent angles.

There are four pair of adjacent angles;

$\angle x$ and

$\angle y, \angle x$ and

$\angle y + \angle z, \angle y$ and

$\angle z , \angle z$ and

$\angle x+ \angle y.$

In Fig. $P,Q$ and $R$ are collinear points and $TQ \bot PR,$
Name:
Four pair of adjacent angles.

There are four pairs of adjacent angles:

$\angle PQT$ and

$\angle TQS, \angle TQS$ and

$\angle SQR, \angle PQT$ and

$\angle TOR$ , and

$\angle PQS$ and

$\angle SQR.$

In Fig., are the above pair of angle adjacent? Justify your answer.

Yes,

$a$ and

$b$ are the adjacent angles because they have a common arm.

In the figure, $OR \bot \ OP$ .
Name all the pairs of complementary angles.

Complementary angles are those angles which adds to

$90^o$ .

Here,

$\angle x$ and

$\angle y$ are complementary angles.
If

$\angle x= \angle y= \angle z$ , then the pairs of complementary angles are;

$\angle x$ and

$\angle y, \angle y$ and

$\angle z$ &

$\angle z$ and

$\angle x$ .

Iron roads $a,\ b,\ c,\ d,\ e$ and $f$ are making a design in a bridge as shown in Fig., in which a $\parallel b,\ c \parallel d,\ e \parallel \ f$ . Find the marked angles between $b$ and $c$

Let the angle between

$b$ and

$c$ is

$1$ , and that is

$30$ degrees by using the property of vertically opposite angles.

1669970_1791102_ans_fbf17c1c99f446a69a8ad07d2b18c6e5.png

In Fig., are the above pair of angle adjacent? Justify your answer.

No,

$a$ and

$b$ don't have a common vertex. Hence, they are not adjacent angles.

What is the type of other angle of a linear pair if one of its angle is right?

In linear pair, if one angle is a right angle then the other will be also a right angle.

Write the complement angle of: $x^{o}$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Given, the angle is

$x^{o}$ .
Then, the complement angle of

$x^{o}$ is,

$=90^{o}-x^{o}$

$=(90-x)^{o}$ .

Therefore, the complement angle of

$x^{o}$ is

$(90-x)^{o}$ .

In Fig., are the above pair of angle adjacent? Justify your answer.

No,

$a$ and

$b$ don't have a common arm. Hence, they are not adjacent angles.

In Fig. write all the pairs of supplementary angles.

They are six pairs of supplementary angles are formed:

$\angle 1$ and

$\angle 8, \angle 2$ and

$\angle 7, \angle 3$ and

$\angle 4, \angle 4$ and

$\angle 5, \angle 5$ and

$\angle 6,$ and

$\angle 3$ and

$\angle 6$ .

In Fig., $OB$ is perpendicular to $OA$ and $\angle BOC=49^o$ . Find $\angle AOD.$

Given,

$\angle AOB=90^o \ \ [OB \bot OA]$

$\Rightarrow \angle BOC + \angle COA=90^o -49^o [\because \angle BOC =49^o]$

$\Rightarrow \angle COA =41^o ......(1)$

Now,

$CD$ is a straight line.

$\therefore \angle AOD + \angle AOC=180^o$ [Linear pair]

$\Rightarrow \angle AOD=180^o-41^o$

$\Rightarrow \angle AOD=139^o$

Write the supplement angle of:
$49^{o}$

The supplement angle of

$49^{o}$ is,

$=180^{o}-49^{o}$

$=131^{o}$
Hence, the supplement angle of

$49^{o}$ is

$131^{o}$

Write the complement angle of: $45^{o}$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then, the complement angle of

$45^{o}$ is,

$=90^{o}-45^{o}$

$=45^{o}$ .

Therefore, the complement angle of

$45^{o}$ is

$45^{o}$ .

In Fig., are the above pair of angle adjacent? Justify your answer.

No,

$a$ and

$b$ are not adjacent angles because their arm is not common and lies on the same side of the common arm.

Write the supplement angle of:
$111^{o}$

The supplement angle of

$111^{o}$ is,

$=180^{o}-111^{o}$

$=69^{o}$
Hence, the supplement angle of

$111^{o}$ is

$69^{o}$

Prove that the bisectors of two adjacent supplementary angles include a right angle.

$\textbf{Step 1: Use Sum of angle of straight line }\boldsymbol{180^{\circ}}\textbf{ to prove angle including bisectors is}\boldsymbol{90^{\circ}}$

$\text{Given,}$

$\angle ACD$ and

$\angle BCD$ are linear pairs.

$CE$ and

$CF$ bisect

$\angle ACD$ and

$\angle BCD$ respectively.

$\text{To prove: }\angle ECF=90^{\circ}$

$\text{Proof:}$

$\text{As we know that angle on a straight line is }180^{\circ}$

$\therefore\angle ACD+\angle BCD=180^{\circ}$

$\Rightarrow 2\angle ECD+2\angle DCF=180^0$

$[\because \boldsymbol{CE}$

$\textbf{and}$

$\boldsymbol{CF}$

$\textbf{bisects}$

$\boldsymbol{\angle ACD}$

$\textbf{and}$

$\boldsymbol{\angle BCD}$

$\textbf{respectively}]$

$\angle ECD+\angle DCF=\dfrac{180^{\circ}}{2}=90^{\circ}$

$\therefore\angle ECF=90^0$

$\textbf{Hence, Proved.}$

The larger of two supplementary angles exceeds thrice the smaller by 20 degrees.
Find them.

$\textbf{Step 1 : Finding of smaller angle.}$

$\text{Let , smaller angle = x and}$

$\text{ larger angle = y }$

According to the question,

$\Rightarrow y=3x + 20$

$\because x+y=180^{\circ}$ [ By linear pair ]

$\Rightarrow x+3x +20=180^{\circ}$

$\Rightarrow x=40^{\circ}$

$\textbf{Step 2: Finding of larger angle}$

$\text{Put value of x in y we get,}$

$\Rightarrow y=3(40)+20$

$=140^{\circ}$

$\textbf{Hence, the two supplementary angles are}$

$\mathbf{ 140^{\circ} and 40^{\circ}}$

2160544_1988068_ans_22d0eee987804ec8b39f159f20ea275a.PNG

Write the supplement angle of:
$20^{o}+y^{o}$

The supplement angle of

$20^{o}+y^{o}$ is,

$=180^{o}-(20^{o}+y^{o})$

$=180^{o}-20^{o}-y^{o}$

$=160^{o}-y^{o}$

$=(160-y)^{o}$
Hence, the supplement angle of

$20^{o}+y^{o}$ is

$(160-y)^{o}$

For an angle $x^{o}$ , find:
the supplementary angle

Supplementary angle of

$x^{o}$ is,

$=(180^{o}-x)$

Write the supplement angle of:
$(x-30)^{o}$

The supplement angle of

$(x-30)^{o}$ is,

$=180^{o}-(x-30)^{o}$

$=180^{o}-x^{o}+30^{o}$

$=210^{o}-x^{o}$

$=(210-x)^{o}$
Hence, the supplement angle of

$(x-30)^{o}$ is

$(210-x)^{o}$

Figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

From given figure,

$\angle AOB + \angle BOC = 97^\circ + 83^\circ = 180^\circ$

The sum of adjacent angles AOB and BOC is

$180^\circ$

Hence, the exterior arms OA and OC are in the same straight line

If true then enter $1$ and if false then enter $0$
Can two obtuse angles be complement to each other?

We know, obtuse angle is the angle which is greater than

$\displaystyle { 90 }^{ o }$ and less than

$\displaystyle { 180 }^{ o }$ .

Two obtuse angles can never be complementary to each other since obtuse angles are itself greater than

$90$ degrees, so the sum of two such angles will always yield a number greater than

$180$ degrees.

E.g.: Let us suppose two obtuse angles are

$100^o$ &

$110^o$ .
Here, the sum of these two obtuse angles will give us

$210^o$ .
Since, for being complementary, sum should be equal to

$90^o$ ,
we can say that no pairs of obtuse angles are complementary.
Hence, the statement is false and

$0$ is the answer.

If true then enter $1$ and if false then enter $0$
Can an acute angle be adjacent to an obtuse angle?

A linear pair is a pair of angles that are adjacent and form a line,

$180^°$ . Unless the two angles are both right angles, then a linear pair is always an acute angle adjacent to an obtuse angle. The diagram is as shown in the above image.

Also, adjacent angles are angles that share a common vertex and edge but do not share any interior points. In other words, they are angles that are side by side, or adjacent, sharing an "arm". Adjacent angles which sum to a right angle, straight angle or full angle are special and are respectively called complementary, supplementary and explementary angles.

Hence, an acute angle is adjacent to an obtuse angle.

1482814_171925_ans_b4687a5340b54acaac97f14e39750f12.png

If true then enter $1$ and if false then enter $0$
Can two obtuse angles be supplementary?

No.

Two obtuse angles can never be supplementary.
Since

$90$ degrees<Obtuse angles<

$180$ degrees,
So sum of two such angles will always greater than 180 degrees.

If true then enter $1$ and if false then enter $0$
Can two right angles be complement to each other?

We know, right angle is the angle equal to

$\displaystyle { 90 }^{ o }$ .

Two right angles can never be a complement to each other.
Since the right angle means

$90$ degrees, so the sum of two right angles will always be

$180$ degrees.

E.g.: Let us suppose two obtuse angles are

$90^o$ &

$90^o$ .
Here, the sum of these two obtuse angles will give us

$180^o$ .
Since, for being complementary, sum should be equal to

$90^o$ ,
we can say that no pairs of right angles are complementary.
Hence, the statement is false and

$0$ is the answer.

If true then enter $1$ and if false then enter $0$
Can two acute angles be supplementary?

No.

Two acute angles can never be supplementary.
Since

$0$ degrees<Acute angles <

$90$ degrees,
So the Sum of two such angles will always be less than 180 degrees.

If true then enter $1$ and if false then enter $0$
Can two obtuse angles form a linear pair?

No.

Two obtuse angles can never form a linear pair.
For two angles to form a linear pair, they need to give a sum of 180 degrees.
But obtuse angles are greater than 90 degrees, so two obtuse angles will always give a sum greater than 180 degrees.

The adjacent angles on a straight line are $(5x-6)^o$ and $7(x+6)^o$ . Find the value of $x$ .

Given that,

${ \left( 5x-6 \right) }^{ o }$ and

${ 7\left( x+6 \right) }^{ o }$ are adjacent angles on a straight line.

Since the sum of adjacent angles on a straight line

$={ 180 }^{ o }$

We can write,,

${ \left( 5x-6 \right) }^{ o }+{ 7\left( x+6 \right) }^{ o }={ 180 }^{ o }$

$\Rightarrow 5x-6+7x+42=180$

$\Rightarrow 12x=180-36$

$\Rightarrow x=\cfrac { 144 }{ 12 }$

$\Rightarrow x=12$

The adjacent angles are as follows:

${ \left( 5x-6 \right) }^{ o }=(5\times 12-6)^o=(60-6)^o={ 54 }^{ o }$

And,

${ 7\left( x+6 \right) }^{ o }=7(12+6)^o=7\times 18^o={ 126 }^{ o }$

The measures of two adjacent angles on a straight line are $x^o$ and $(2x-27)^o$ . Find the smallest angle.

Given:

${ x }^{ o }$ and

${ \left( 2x-27 \right) }^{ o }$ are adjacent angles on a straight line.

Now,

${ x }^{ o }+{ \left( 2x-27 \right) }^{ o }={ 180 }^{ o }$ ...(Since adjacent angles on a straight line

$={ 180 }^{ o }$ )

$\Rightarrow 3x=180+27$

$\Rightarrow 3x=207$

$\Rightarrow x=\cfrac { 207 }{ 3 }$

$\Rightarrow x=69$

The ajacent angles are as follows:

${ x }^{ o }={ 69 }^{ o }=$ Smallest angle

And,

${ 2x-27 }^{ o }=2\times 69-27=138-27={ 111 }^{ o }$

The side EF, FD and DE of a triangle DEF are produced in order forming three exterior angles DFP, EDQ and FER respectively. Prove that $\angle DFP+\angle EDQ+\angle FER=360^o$ .

For triangle

$DEF,$ let the angles are

$x,y$ and

$z$

then

$x+y +z =180^{o}$

$\angle DFP = 180^o -x$

$\angle EDQ = 180^o -y$

$\angle FER = 180^o -z$
Add these three equations, we get

$\angle DFP+\angle EDQ+\angle FER = 540^o - (x+y+z)$

$= 540^o - 180^o$

$= 360^o$

Hence,

$\angle DFP+\angle EDQ+\angle FER =360^o$

Prove that if the sum of two adjacent angles is $180^o$ , then the non-common arms are two opposite rays.

Given: Two adjacent angles are

$\angle AOC$ and

$\angle BOC$ and

$\angle AOC+\angle BOC=180^o$
To prove: OA and OB are two opposite rays.
Construction: Let OA and OB are not two opposite rays.
Then, draw a ray OE opposite to OA such that AOE is a straight line.
Proof:

$\angle AOC+\angle BOC=180^o$ .....(1)

$\angle AOC+\angle COE=180^o$ ....(2) (Linear pair)

$\angle AOC+\angle EOC=\angle AOC+\angle BOC$ (From equation 1 and 2)

$\therefore \angle EOC=\angle BOC$
This is possible only if OE and OB coincide. Hence, OA and OB are two opposite rays.

In Fig. 10.16 $\displaystyle \angle OAB=30^{\circ}\,and\,\angle OCB=57^{\circ}$ Find $\displaystyle \angle BOC\,and\,\angle AOC$

$\Rightarrow$

$\angle OAB=30^o$ and

$\angle OCB=57^o$ .

$\triangle AOB,$

$\Rightarrow$

$AO=OB$ [ Radius of a circle ]

$\Rightarrow$

$\angle OBA=\angle BAO=30^o$ [ Angles opposite to equal sides are equal ]
In

$\triangle AOB,$

$\Rightarrow$

$\angle AOB+\angle OBA+\angle BAO=180^o.$

$\therefore$

$\angle AOB+30^o+30^o=180^o$

$\therefore$

$\angle AOB=180^o-30^o-30^o$

$\therefore$

$\angle AOB=120^o$ ----- ( 1 )

Now, in

$\triangle OCB,$

$\Rightarrow$

$OC=OB$ [ Radius of a circle ]

$\Rightarrow$

$\angle OBC=\angle OCB=57^o.$ [ Angles opposite to equal sides are equal ]

$\triangle OCB,$

$\Rightarrow$

$\angle BOC+\angle OCB+\angle OBC=180^o.$

$\Rightarrow$

$\angle BOC+57^o+57^o=180^o$

$\Rightarrow$

$\angle BOC=180^o-114^o$

$\therefore$

$\angle BOC=66^o$ ------ ( 2 )

$\Rightarrow$

$\angle AOB=120^o$ [ From ( 1 ) ]

$\Rightarrow$

$\angle AOC+\angle COB=120^o.$

$\Rightarrow$

$\angle AOC+66^o=120^o$

$\Rightarrow$

$\angle AOC=120^o-66^o$

$\therefore$

$\angle AOC=54^o$

In the figure, sides $QP$ and $RQ$ of $\Delta PQR$ are produced to points $S$ and $T$ respectively. If $\angle SPR={ 135 }^{o }$ and $\angle PQT={ 110 }^{ o}$ , find $\angle PRQ$ .

Given,

$\angle SPR = 135^{\circ} ,\angle PQT = 110^{\circ}$
According to the question

$\angle SPR +\angle QPR = 180^{\circ}$ (

$SQ$ is a straight line.)

$\Rightarrow 135^{\circ} +\angle QPR = 180^{\circ}$

$\Rightarrow \angle QPR = 45^{\circ}$

Also

$\angle PQT +\angle PQR = 180^{\circ}$ (

$TR$ is a straight line.)

$\Rightarrow 110^{\circ} +\angle PQR = 180^{\circ}$

$\Rightarrow \angle PQR = 70^{\circ}$

Now,

$\angle PQR +\angle QPR + \angle PRQ = 180^{\circ}$ (Sum of the interior angles of the triangle.)

$\angle 70^{\circ} + 45^{\circ} + \angle PRQ = 180^{\circ}$

$\Rightarrow 115^{\circ} + \angle PRQ = 180^{\circ}$

$\Rightarrow \angle PRQ = 65^{\circ}$

It is given that $\angle XYZ={ 64 }^{ }$ and $XY$ is produced to point $P$ . Draw a figure from the given information. If ray $YQ$ bisects $\angle ZYP$ , find $\angle XYQ$ and reflex $\angle QYP$ .

Given,

$\angle XYZ = 64^{\circ}$

$YQ$ bisects

$\angle ZYP$

$\angle XYZ +\angle ZYP = 180^{\circ}$ (Linear Pair)

$\Rightarrow 64^{\circ} +\angle ZYP = 180^{\circ}$

$\angle ZYP = 116^{\circ}$

Also,

$\angle ZYP = \angle ZYQ + \angle QYP$

$\angle ZYQ = \angle QYP$ (

$YQ$ bisects

$\angle ZYP$ )

$\Rightarrow \angle ZYP = 2\angle ZYQ$

$\Rightarrow 2\angle ZYQ = 116^{\circ}$

$\Rightarrow \angle ZYQ = 58^{\circ} = \angle QYP$

Now,

$\angle XYQ = \angle XYZ + \angle ZYQ$

$\Rightarrow \angle XYQ = 64^{\circ} + 58^{\circ}$

$\Rightarrow \angle XYQ = 122^{\circ}$

Also, reflex

$\angle QYP = 180^{\circ} + \angle XYQ$

$\angle QYP = 180^{\circ} + 122^{\circ}$

$\Rightarrow \angle QYP = 302^{\circ}$

In the figure, lines $XY$ and $MN$ intersect at $O$ . If $\angle POY={ 90 }^{o }$ and $a : b = 2 : 3$ , find $c$ .

Given,

$\angle POY = 90^{\circ}$ and

$a : b = 2 : 3$
According to the question,

$\angle POY + a + b = 180^{\circ}$

$\Rightarrow 90^{\circ}+ a + b = 180^{\circ}$

$\Rightarrow a + b = 90^{\circ}$
Let

$a$ be

$2x$ , then will be

$3x$

$\Rightarrow 2x + 3x = 90^{\circ}$

$\Rightarrow 5x = 90^{\circ}$

$\Rightarrow x = 18^{\circ}$

$\therefore a = 2\times 18 = 36^{\circ}$

$b = 3×18= 54^{\circ}$

$b + c = 180^{\circ}$

$\mid$ Linear Pair

$\Rightarrow 54^{\circ}+ c = 180^{\circ}$

$\Rightarrow c = 126^{\circ}$

The line $d$ is parallel to $e$ , as shown in $y$ .

In the figure,

$y$ is the exterior angle of the triangle shown, having two of its opposite interior angles as

$90^{\circ}$ and

$58^{\circ}$ .

We know, by exterior angle property, an exterior angle of a triangle is the sum of two of its opposite interior angles.

$\therefore y = 90^{\circ} + 58^{\circ} = 148^{\circ}$

Hence, the value of

$y$ is

$148^{\circ}$ .

Which pair of angles are supplementary in the following diagram? Are there supplementary rays?

The total sum of angles

$=360^o$

$\Rightarrow 360^o=50^o+120^o+130^o+\angle AOD$

$\therefore \angle AOD=60^o$

None of the pairs are supplementary. For a pair of angle to be supplementary, their sum should be

$180^o$ .

Here,

$\angle BOC$ and

$\angle COD= 120^o+130^o=250^o \neq 180^o$

$\angle COD$ and

$\angle AOD= 130^o+60^o=190^o \neq 180^o$

$\angle DOA$ and

$\angle AOB= 60^o+50^o=110^o \neq 180^o$

$\angle BOA$ and

$\angle BOC= 50^o+120^o=170^o \neq 180^o$

Two complementary angles differ by ${12}^{o}$ , find the angles?

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

Given, two complementary angles differ by $20^o$ .

$\Rightarrow x-(90^o -x)=12^o$

$\Rightarrow x-90^o +x=12^o$

$\Rightarrow \ 2x=12^o+90^o$

$\Rightarrow \ x=\dfrac {102^o}{2}$

$\Rightarrow \ x=51^o$ .

Therefore, complementary of $x$ ,

$=90^o -x$

$=90^o -51^o$

$=39^o$ .

Hence, $51^o$ and $39^o$ are the complementary angles.

Draw diagrams illustrating each of the following situation:
(a) Three straight lines which do not pass through a fixed point.
(b) A point and rays emanating from that point such that the angle between any two adjacent rays is an acute angle.
(c) Two angles which are not adjacent angles, but still supplementary.
(d) Three points in the plane which are equidistant from each other.

$\angle SPQ=$ Acute Angle

$\angle QPR=$ Acute Angle

$\angle RPT=$ Acute Angle

1174349_558674_ans_df527e792ca04041b481dbb2d4fc491a.PNG

Find the value of $x$ in the following diagram:

$\angle AOC+ \angle BOC=180^{\circ}$

$[\because$ linear pair of angles

$]$

$\implies (x-y)+(x+y)=180^{\circ}$

$\Rightarrow 2x=180^{\circ}$

$\therefore x=90^{\circ}$

Statement
$\angle 1$ is supplement to $\angle 8$ .

Provide appropriate reason for the same if $l$ is parallel to $m$ and a transversal intersects these parallel lines.

Reason

$:$

$\angle 1=\angle 5$ (Corresponding angles as

$l||m$ )

$\angle 5+ \angle 8={ 180 }^{ \circ }$ (Supplementary angles made on a line made at a point)

Put

$\angle 5 = \angle 1$

$\angle 1+\angle 8 ={ 180 }^{ \circ }$

$\angle1$ is supplement to

$\angle8$

In the figure given above the sides, AB and AC of $\triangle$ ABC are produced to points E and D respectively. If bisectors BO and CO of $\angle CBE$ and $\angle BCD$ respectively meet at point O, then prove that $\angle BOC = 90^o - \dfrac{1}{2} \angle BAC$ .

$\angle CBE + y^o = 180^o$ .........(a) {Linear Pair}

$\angle BCD + z^o = 180^o$ .........(b) {Linear Pair}

Ray BO is the bisector of

$\angle CBE$ .

Therefore,

$\angle CBO = \dfrac{1}{2} \angle CBE$

$= \dfrac{1}{2} (180^o - y^o)$ {from (a)}

$= 90^o - \dfrac{y^o}{2}$ ...... (1)

Similarly, ray CO is the bisector of

$\angle BCD$ .

Therefore,

$\angle BCO = \dfrac{1}{2} \angle BCD$

$= \dfrac{1}{2} (180^o - z^o)$ {from (b)}

$= 90^o - \dfrac{z^o}{2}$ ...... (2)

$\triangle BOC,$

$\angle BOC + \angle BCO + \angle CBO = 180^o$ ...... (3)

Substituting (1) and (2) in (3), we get,

$\angle BOC + 90^o - \dfrac{z^o}{2} + 90^o - \dfrac{y^o}{2} = 180^o$

So,

$\angle BOC = \dfrac{z^o}{2} + \dfrac{y^o}{2}$

or,

$\angle BOC = \dfrac{1}{2} (y^o + z^o)$ ...... (4)

But,

$x^o + y^o + z^o = 180^o$ (Angle sum property of a triangle)

Therefore,

$y^o + z^o = 180^o - x^o$

Therefore, (4) becomes

$\angle BOC = \dfrac{1}{2} (180^o - x)^o$

$= 90^o - \dfrac{x}{2}$

$= 90^o - \dfrac{1}{2} \angle BAC$

1039464_570819_ans_52575dbe59fd45e19dc09313b681b890.jpg

$\triangle{PQR}$ is an isosceles triangle with $PQ=PR$ , $QP$ is produced to $S$ and $PT$ bisects the extension angle $2{ x }^{ o }$ . Prove that $\angle Q={ x }^{ o }$ and hence prove that $PT\parallel QR$ .

Given:

$\triangle {PQR}$ is an isosceles triangle with

$PQ=PR$
Proof:

$PT$ bisects exterior angle

$\angle {SPR}$ and therefore

$\angle SPT=\angle TPR={ x }^{ o }$

$\therefore$

$\angle Q=\angle R$ (Property of an isosceles triangle)
also we know that in any triangle,
exterior angle

$=$ sum of the interior opposite angles.

$\therefore$ In

$\triangle {PQR}$ , Exterior angle

$\angle SPR=\angle PQR+\angle PRQ$

$2{ x }^{ o }=\angle Q+\angle R$

$=\angle Q+\angle Q$

$2{ x }^{ o }=2\angle Q$

${ x }^{ o }=\angle Q$
To prove:

$PT\parallel QR$
Lines

$PT$ and

$QR$ are cut by the transversal

$SQ$ . We have

$\angle SPT={ x }^{ o }$
Hence,

$\angle SPT$ and

$\angle PQR$ are corresponding angles:

$PT\parallel QR$

In the given figure $AB || CD$ and $CD || EF$ . Also $EA \perp AB$ . If $\angle BEF = 55^\circ$ find the values of $x, y$ and $z$ .

Extend BE to G.

Now

$\angle GEF = 180^o - 55^o$ Linear pair

$= 125^o$

Also

$\angle GEF = x = y = 125^o$ Corresponding angles

Now,

$z = 90^o - 55^o$ Right angle

$= 35^o$

1034830_570699_ans_fb85020011d641ab84ca87baaaaa7e19.jpg

Using information given in the adjacent figure, find the values of $x$ and $y.$

Side BC of

$\triangle$ ABC has been produced to D.

Exterior

$\angle ACD = \angle ABC + \angle BAC$

$\therefore 100^o = 65^o + x^o$

$\Rightarrow x^o = (100^o - 65^o) = 35^o$

$\therefore \angle CAD = \angle BAC = 35^o$

$\triangle$ ACD, we have:

$\angle CAD + \angle ACD + \angle CDA = 180^o$ (Angle sum property of triangle)

$\Rightarrow 35^o + 100^o + y^o = 180^o$

$\Rightarrow 135^o + y^o = 180^o$

$\Rightarrow y^o = (180^o - 135^o) = 45^o$

Hence,

$x = 35^o, y = 45^o$ .

1034838_570809_ans_953f99ad711944e595362f49080f1dc8.jpg

In the adjacent figure PQ and RS are two mirrors placed parallel to each other. An incident ray $\overrightarrow{AB}$ strikes the mirror PQ at B, the reflected ray moves along the path $\overrightarrow{BC}$ and strikes the mirror RS at C and again reflected back along $\overrightarrow{CD}$ . Prove that AB || CD.

Since

$BE$ and

$FC$ are normal to

$PQ$ and

$RS$ respectively, therefore,

$BE||FC$

Let,

$\angle ABE=\angle EBC=x(PQ$ is a mirror so angle of incidence is equal to angle of reflection)

$\angle FCD=\angle BCF=y$ (RS is a mirror so angle of incidence is equal to angle of reflection)

Now considering BE and FC, taking BC as transversal,

$\angle EBC=\angle BCF........(i)$ (alternate interior angle)

i.e.

$x=y$

i.e.

$\angle ABE=\angle FCD......(ii)$

adding equation (i) and (ii)

$\angle EBC+\angle ABE=\angle BCF+\angle FCD$

$\angle ABC=\angle BCD$

Now if we take line

$AB$ and

$CD$ in consideration, alternate interior angle that are

$\angle ABC$ and

$\angle BCD$ are equal.

Therefore,

$AB||CD$

1073575_570720_ans_e8cbf02f6ed04055bb723aab23834030.png

In the given above figure, AB || QR, $\angle BAQ = 142^o$ and $\angle ABP = 100^o$ . Find $\angle APB$ .

$\angle BAQ + \angle BAP = 180^o$ {Linear pair or angles on the straight line}

$142^o + \angle BAP = 180^o$

$\angle BAP = 180^o-142^o = 38^o$

Also,

$\angle APB +\angle ABP +\angle BAP = 180^o$ {angle sum property of a triangle ABP}

$\angle APB +100^o +38^o = 180^o$

$\angle APB = 180^o-100^o-38^o$

$\angle APB = 42^o$

Find the supplementary angle of the given angle(in degrees).
$105^0$

Supplementary angles are two angles with a sum of

$180^o.$

One supplement angle is

$105^o.$

$\Rightarrow$ Other supplement angle

$= 180^o-105^o$

$=75^o$

Hence, the required supplementary angle is

$75^o$ .

Look at the pictures. Mark the angles and colour them as indicated below.
(a) If the angle is right angle then mark it in blue.
(b) If an angle is less than a right angle then mark it in red.
(c) If an angle is more than a right angle then mark it in black.

1369154_694933_ans_635cb30646944d7b93772db05225948e.png

Find the number of pairs of adjacent angles in the figure .

1308800_702897_ans_0160a028bf8e4654905ad01ba5326178.jpg

How many of the following are adjacent angles ?

1308794_702893_ans_6dc64b2afcfd4791bca5fbc6a2641500.jpg

What is the measure of complementary angle of $20^{\circ}$ ?
(1) $60^{\circ}$
(2) $70^{\circ}$
(3) $80^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $20^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-20^o$ $=70^o.$

$\therefore$ Measure of a complementary angle of $20^{o}=70^o$ .

Therefore, $(2)$ $70^o$ is correct.

What is the measure of complementary angle of $55^{\circ}$ ?
(1) $25^{\circ}$
(2) $15^{\circ}$
(3) $35^{\circ}$

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $55^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-55^o$ $=35^o.$

$\therefore$ Measure of a complementary angle of $55^{o}=35^o$ .

Therefore, $(3)$ $35^o$ is correct.

What is the measure of complementary angle of $83^{\circ}$ ?
(1) $7^{\circ}$
(2) $17^{\circ}$
(3) $27^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $83^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-83^o$ $=7^o.$

$\therefore$ Measure of a complementary angle of $83^{o}=7^o$ .

Therefore, $(1)$ $7^o$ is correct.

Write the measure of supplementary angle of the given angles:
$145^{\circ}$ .

$180^{\circ} - 145^{\circ} = 35^{\circ}$

Thus, Supplementary angle of given angle is

$35^{\circ}$ .

Measure of one angle of linear pair is given as $50^{\circ}.$ Find the measure of another angle.

We know that sum of angles in linear pair is

$180^\circ$

Measure of another angle

$=180^{\circ} - 50^{\circ}$

$= 130^{\circ}$

Thus, another angle is

$130^{\circ}$

Two angles are supplementary angles. If measure of one angle is $66^{\circ}$ , then find the measure of another angle.

Two angles are supplementary angles

$\therefore \angle A+\angle B=180^\circ$

$\therefore 180^{\circ} - 66^{\circ} = 114^{\circ}$

Thus, the supplementary angle of given angle is

$114^{\circ}$

Find complementary angle of $23^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $23^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-23^o$ $=67^o.$

$\therefore$ Measure of a complementary angle of $23^{o}=67^o$ .

One angle of linear pair is a right angle, then find the measure of another angle.

$180^{\circ} - 90^{\circ} = 90^{\circ}$

Thus, another angle is

$90^{\circ}$

Measure of one angle of linear pair is given. Find the measure of another angle:
$85^{\circ}$ .

$180^{\circ} - 85^{\circ} = 95^{\circ}$

Thus, another angle is

$85^{\circ}$

$85^{\circ}$ .

From the given figure write two pairs of supplementary angles:

$1.$

$\angle EFI$ and

$\angle GFI$ are supplementary angles because sum of these two angles is

$180^{\circ}$ .

$2.$

$\angle EFH$ and

$\angle GFH$ are also supplementary angles because sum of these two angles is

$180^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$107^{\circ}$ .

$180^{\circ} - 107^{\circ} = 73^{\circ}$

Thus another angle is

$73^{\circ}$

Measure of one angle of linear pair is $82^{\circ}$ , find the measure of another angle.

$180^{\circ} - 82^{\circ} = 98^{\circ}$

Thus, another angle is

$98^{\circ}$

Measure of one angle of linear pair is given. Find the measure of another angle:
$155^{\circ}$ .

$180^{\circ} - 155^{\circ} = 25^{\circ}$

Thus, another angle is

$25^{\circ}$

Measure of one angle of linear pair is $108^{\circ}$ , then find the measure of another angle.

$180^{\circ} - 108^{\circ} = 72^{\circ}$

Thus, measure of another angle is

$72^{\circ}$

Write all pairs of angles of linear pair.

$\angle XMQ$ and

$\angle QMY, \angle QMY$ and

$\angle YMP, \angle YMP$ and

$\angle PMX, \angle PMX$ and

$\angle XMQ$

$\overleftrightarrow{AB}$ || $\overleftrightarrow{CD}$ in the following figure. $\overleftrightarrow{XY}$ is the transversal. If $m \angle PQA = 140^o$ , find the measure of remaining angles:

$m \angle XPC = 140^o$ ,

$m \angle DPQ = 140^o$ ,

$m \angle BQY = 140^o$ ,

$m \angle AQY = 40^o$ ,

$m \angle PQB = 40^o$ ,

$m \angle CPQ = 40^o$ ,

$m \angle XPD = 40^o$ .

In the following figure, $ABCD$ and $PORS$ are two parallelograms such that $\angle D=120^{o}$ and $\angle Q=70^{o}$ . Find the value of $x$

Since

$PQRS$ is parallelogram

$\angle S=\angle Q$ (opposite angles of a parallelogram)

$\angle S=70^{\circ}$

Now,

$ABCD$ is a parallelogram

$\angle D+\angle C=180^{\circ}$

${120^\circ } + \angle C = {180^\circ }$

$\angle C = {60^\circ }$

Therefore,

$\angle S +\angle C+x=180^{\circ}$

$70^{\circ}+60^{\circ}+x=180^{\circ}$

$x=50^{\circ}$

Find $\angle a,\angle b,\angle c, \angle d$ and $\angle e$ in the figure? Give reasons.

$\angle a=50^{\circ}$ (alternate interior angles)

$\angle b=50^{\circ}$ (alternate interior angles)

$\angle c=50^{\circ}$ (alternate interior angles)

$\angle d=50^{\circ}$ (alternate interior angles)

$\angle e=50^{\circ}$ (alternate interior angles)

In $\Delta PQR,\angle P = 2\angle Q$ and $\angle R = 3\angle Q$ , calculate the angles of $\Delta PQR$

Given,

$\angle P=2\angle Q$

$\angle R=3\angle Q$

We know that in

$\triangle PQR$

$\angle P+\angle Q+\angle R=180$

$2\angle Q+\angle Q+3\angle Q=180$

$6\angle Q=180$

$\angle Q=\dfrac{180}{6}=30^0$

$\angle P=2\angle Q=60^0$

$\angle R=3\angle Q=90^0$

Observe the figure and find $x$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $40^o$ and $(6x+2)^o$ .

Then, their sum: $40^o+(6x+2)^o=90^o$ .

$\implies$

$40^o+6x+2^o=90^o$

$\implies$

$42^o+6x=90^o$

$\implies$

$6x=90^o-42^o$

$\implies$

$6x=48^o$

$\implies$

$x=\dfrac{48^o}{6}$

$\implies$

$x=8^o$ .

In fig. if $AB || CD$ then find the value of $x$ .

From the figure we have,

$\angle EPB=180^o-(5x-20^o)$ ......(1)[ Since

$\angle BPE+\angle QPB=180^o$ ]

Again

$\angle PQD=\angle QPB$ [ Similar angles]

or,

$180^o-(5x-20^o)=2x-10^o$

or,

$7x=210^o$

or,

$x=\dfrac{210^o}{7}=30^o$ .

In parallelogram $ABCD$ two points $P$ and $Q$ are taken on diagonal $BD$ such that $DP=BQ (see fig. 8.20)$ .
Show that: $AP = CQ$

Given

$ABCD$ is parallelogram and given

$BD$ is diagonal such that

$DP= BQ$

$\Rightarrow$ Now

$AD \parallel BC$ [

$AD$ is parallel to

$BC$ ]

$BD$ is transversal i.e., diagonal

$\triangle APD$ &

$\triangle CQB$

$AD= CB$

$\angle ADP= \angle CBQ$ [Diagonals bisect the angles ]

$DP= BQ$

$\therefore \triangle APD \cong \triangle CQB$ [

$SAS$ concurrency ]

$\therefore AP= CQ$

From the figure given above, find the values of angles $x$ , $y$ and $z$ .

We know that,

$40^o + x + 25^o = 180^o$ [Angle on straight line]

$\Rightarrow 65^o + x = 180^o$

$\Rightarrow x = 180^o - 65^o = 115^o$

Similarly,

$40^o + y = 180^o$ [Linear pair]

$\Rightarrow y = 180^o - 40^o = 140^o$ ....(1)

Also

$y +z = 180^o$ [Linear pair]

$\Rightarrow 140^o + z = 180^o$ [From eq. (1)]

$\Rightarrow z = 180^o - 140^o = 40^o$

Thus,

$x = 115^o, \, y = 140^o$ and

$z = 40^o$

1081520_1048128_ans_dad973125e72485b8b213cf3f6983194.png

In figure, $m||n$ and $p||q$ . If $\angle 1=75^{o}$ , prove that $\angle 2=\angle 1+\dfrac {1}{3}$ of a right angle.

Given:

$m\parallel n$ and

$p\parallel q$

$\Rightarrow \angle 3 = \angle 1$ (corresponding angles)

$\Rightarrow \angle 3 = 75^\circ$

Since

$m\parallel n$ ,

Now ,

$\angle 2 + \angle 3 = 180^\circ$ (cointerior angles)

$\Rightarrow \angle 2 + 75^\circ = 180^\circ$

$\Rightarrow \angle 2 = 105^\circ$

RHS

$= \angle 1 + \frac{1}{3} \times 90^\circ$

$= 75^\circ + 30^\circ$

$= 105^\circ$

$= \angle 2$

$= LHS$

Hence proved

If two lines intersect, prove that the vertically opposite angles are equal.

${\textbf{Step - 1: Find pair of Opposite Angles}}$

${\text{If two lines i}}{\text{.e}}{\text{. AB & CD intersect each other}}{\text{. }}$

${\text{They have two pair of opposite angles}}{\text{. i}}{\text{.e}}{\text{.}}\angle {\text{AOC,}}\angle {\text{DOB,}}\angle {\text{AOB,}}\angle {\text{COB}}{\text{.}}$

$\angle {\text{AOC + }}\angle {\text{AOD = 18}}{{\text{0}}^{\text{o}}}{\text{ }}\left[ {{\text{linear pair}}} \right]{\text{ .......}}\left( {\text{1}} \right)$

$\angle {\text{AOD + }}\angle {\text{BOD = 18}}{{\text{0}}^{\text{o}}}{\text{ }}\left[ {{\text{linear pair}}} \right]{\text{ ..........}}\left( {\text{2}} \right)$

${\textbf{Step -2: prove of vertically opposite angles are equal}}$

${\text{from eq}}{\text{. }}\left( {\text{1}} \right){\text{ & }}\left( {\text{2}} \right)$

$\angle {\text{AOC + }}\angle {\text{AOD = }}\angle {\text{AOD + }}\angle {\text{BOD}}$

$\Rightarrow {\text{}}\angle {\text{AOC = }}\angle {\text{BOD}}$

${\text{Similarly }}\angle {\text{AOD = }}\angle {\text{COB}}$

${\textbf{Final Answer: Hence proved vertically opposite angles are equal}}{\text{.}}$

GOAL is a quadrilateral in which GO $||$ AL. If $\angle$ G $=\angle$ O $=40^o$ . What are the measure of $\angle$ A and $\angle$ L.

1324937_1064994_ans_4e9b6792985d4df18809447c2ff40b71.jpg

An angle is equal to four times its complement. Determine its measure.

Let the measure of the angle is

$x^{\circ}$

Then its complement is

$\left ( 90-x \right )^{\circ}$

Now, the angle is equal to four times its complement.

Then,

$x=4\left ( 90-x \right )$

$\implies$

$x=360^0-4x$

$\implies$

$x+4x=360^0$

$\implies$

$5x=360$

$\implies$

$x=\dfrac{360}{5}=72^{\circ}$ .

Ray $PQ$ and ray $PR$ are perpendicular to each other. Points $B$ and $A$ are in the interior and exterior of $\angle QPR$ respectively. Ray $PB$ and ray $PA$ are perpendicular to each other. Draw a figure all this rays and write-A pair of complementary angles A pair of supplementary anglesA pair of congruent angles.

Ray

$PQ$ and

$PR$ are perpendicular each other. Point

$B$ is the interior and Point

$A$ is the exterior of

$\angle QPR$ .

Ray

$PB$ and

$PA$ are perpendicular to each other.

$(i)$ A pair of Complementary angles:

$\Rightarrow$

$\angle QPB$ and

$\angle BPR$ because their sum is

$90^o$

$\Rightarrow$

$\angle BPR$ and

$\angle RPA$ because their sum is

$90^o$

$(ii)$ A pair of Supplementary angles:

$\Rightarrow$

$\angle QPR$ and

$\angle BPA$ because their sum is

$180^o$

$(iii)$ A pair of congruent angles:

$\Rightarrow$

$\angle BPA\equiv\angle QPR$ [ Each angle is

$90^o$ ]

$\Rightarrow$

$\angle BPA=\angle QPR$

So,

$\angle RPA+\angle BPR=\angle BPR+\angle QPB$ [ Using angle addition property ]

So,

$\angle RPA=\angle QPB$

Hence,

$\angle RPA\equiv \angle QPB$

997864_1060884_ans_10a22522700341a7aca5d90bb4944061.png

In Figure, AB $||$ CD and EF is a transversal. Find the measure of all the other angles that are formed.

$AB||CD$

$EF$ is transversal

$\angle AGE=45^{\circ}\\\angle EGB+45=180\\\implies \angle EGB=135^{\circ}\\\angle AGH=135^{\circ}\\\implies \angle GHC=45\\\quad \angle GHD=135\\\quad \angle DHF=45\\\quad \angle FHC=135\\\quad \angle GHC=45$

Find the value of $x$ in the above figure.

Using the property of linear pair, we have

$x-10^{\circ}+4x-25^{\circ}+x+5^{\circ}={180}^{\circ}$

$\Rightarrow\,6x-30^{\circ}=180^{\circ}$

$\Rightarrow\,6x=180^{\circ}+30^{\circ}=210^{\circ}$

$\Rightarrow\,x=\dfrac{210^{\circ}}{6}=35^{\circ}$

Hence

$x-10^{\circ}=35^{\circ}-10^{\circ}=25^{\circ}$

$4x-25^{\circ}=4\times\,35^{\circ}-25^{\circ}=140^{\circ}-25^{\circ}=115^{\circ}$

and

$x+5^{\circ}=35^{\circ}+5^{\circ}=40^{\circ}$

An angle is $\dfrac {2}{3}$ of its supplementary. What is its magnitude?

$\Rightarrow$ Let one angle be

$x$ .

$\Rightarrow$ Then, other angle will be

$\dfrac{2}{3}x$

According to the question,

$x+\dfrac{2}{3}x=180^o$

$\dfrac{3x+2x}{3}=180^o$

$\dfrac{5x}{3}=180^o$

$5x=540^o$

$\therefore$

$x=108^o$

$\therefore$

$\dfrac 23 x=\ \dfrac{2}{3} \times 108^o=72^o$

$\therefore$ Magnitudes are

$108^o$ and

$72^o$

In the figure, find x.

We are given,

As we know any two parallel axis opposite each other make angle

$\pi (or\ 180^{o})$

So,

$2x + 4x= \pi$

$6x = \pi$

$x =\dfrac{ \pi }{6}$

now,

$\angle ROP= 2x = \dfrac{2 \pi}{6} = \dfrac{\pi}{3} (or\ 60^{o})$

$\angle ROQ= 4x =\dfrac{4 \pi}{6}= \dfrac{2 \pi}{3} (or\ 120^{o})$

So, value of

$x$ is

$\pi/6$ or

$30^{o}$

1424442_1064974_ans_600e90036cb74de29bf3446e5b60a6b4.png

Find the angles in a straight line

Given ratio for the angles are:

$x,2$ , and

$3$

Since they form a straight line, sum of the angles must be 180.

Therefore,

$x+2x+3x=180$

$6x=180$

$\Rightarrow x=30$

Therefore the angles are:

$\angle x=30$

angle

$2 = 2x =60$

angle

$3 = 3x =90$

How many sets of parallel line segments can you find in a cube?

$ABDCHEFGA$ is a cube.

From the fig,The side

$AB$ has

$3$ sets of parallel line segments

$CD,HE,GF=3$ sets

The side

$CD$ has

$3$ sets of parallel line segments

$AB,HE,GF$ in which

$AB\parallel CD$ is repeated.Hence there are

$3-1=2$ sets

The side

$EH$ has

$3$ sets of parallel line segments

$AB,CD,GF$ in which

$AB\parallel CD$ and

$HE\parallel CD$ are repeated.Hence there are

$3-2=1$ set.

Horizontally

$=3+2+1=6$ sets of parallel line segments can be seen in a cube

Consider the side

$AG$ has

$3$ sets of parallel line segments

$BF,DE,CH=3$ sets

The side

$BF$ has

$3$ sets of parallel line segments

$CH,AG,BF$ in which

$BF\parallel AG$ is repeated.Hence there are

$3-1=2$ sets

The side

$CH$ has

$3$ sets of parallel line segments

$AG,BF,DE$ in which

$CH\parallel AG$ and

$DE\parallel BF$ are repeated.Hence there are

$3-2=1$ set.

Vertically

$=3+2+1=6$ sets of parallel line segments can be seen in a cube

Total sets of parallel line segments can be seen in a cube

$=6+6=12$

1509012_1056128_ans_6b8b44b70d9f46739123894d34bd6abc.PNG

Find the value of $x$ :

Consider the given fig.

$(2x+3)+\left( x-6 \right)={{180}^{0}}$ (supplementary angle)

$(2x+3)+\left( x-6 \right)={{180}^{0}}$

$3x=183$

$x=\dfrac{183}{3}={{61}^{0}}$

Hence, this is the answer.

1053539_1085620_ans_0e91b03bbcfc43f5a3b652b92676e719.png

Two complementary angles differ by $10^{o}$ . Find the angles.

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

Given, two complementary angles differ by $20^o$ .

$x+\left( x-{{10}^{o}} \right)={{90}^{o}}$

$2x-{{10}^{o}}={{90}^{o}}$

$2x={{100}^{o}}$

$x={{50}^{o}}$ .

Then, the angles are $x=50^o$ and other angle is $x-{{10}^{o}}={{50}^{o}}-{{10}^{o}}={{40}^{o}}$ .

Two complementary angles differ by ${6^ \circ }$ . Find the measure of the two angles.

Let $x$ be the one angle.

So, the complement of $x=90^{\circ} -x$ .

Given, two complementary angles differ by $6^{\circ}$ .

$\Rightarrow x-(90^{\circ} -x)=6^{\circ}$

$\Rightarrow x-90^{\circ} +x=6^{\circ}$

$\Rightarrow \ 2x=6^{\circ}+90^{\circ}$

$\Rightarrow \ x=\dfrac {96^{\circ}}{2}$

$\Rightarrow \ x=48^{\circ}$ .

Therefore, complementary of $x$ ,

$=90^{\circ} -x$

$=90^{\circ} -48^{\circ}$

$=42^{\circ}$ .

Hence, $48^{\circ}$ and $42^{\circ}$ are the complementary angles.

On the sides, AB and AC of triangle ABC, equilateral triangles ABD and ACE are drawn. Prove that: $\angle {\text{CAD}} = \angle {\text{BAE}}$

Given:

$\triangle{ABD}$ and

$\triangle{ACE}$ are equilateral triangle.

$\therefore$ In

$\triangle{ABD}$

$\angle{BAD}=\angle{ABD}=\angle{BDA}={60}^{\circ}$ .....

$(1)$

$\triangle{ACE}$

$\angle{AEC}=\angle{ACE}=\angle{ACE}={60}^{\circ}$ .....

$(2)$

To prove:

$\angle{CAD}=\angle{BAE}$

Proof:

$\angle{CAD}=\angle{BAC}+\angle{DAB}$

$\Rightarrow \angle{CAD}=\angle{BAC}+{60}^{\circ}$ using

$(1)$

$\Rightarrow \angle{CAD}=\angle{BAC}+\angle{CAE}$ using

$(2)$

$\Rightarrow \angle{CAD}=\angle{BAE}$

Hence proved.

1258825_1078789_ans_c3ba21e176b0400dab5df7076c0b6a88.PNG

In figure, $l || m$ and $p || q$ . Find the values of $x, y, z$ and $t$

Now

Here

$z = 80^{\circ}$ (

$\because$ vertically opposite angles are same)

Now

$z = t$ (

$\because$ corresponding angles are equal)

So,

$t = 80^{\circ}$

Again

$y$ and

$z$ are corresponding angles

so,

$y = z = 80^{\circ}$

Again

$x \ \& \ y$ are also corresponding angles

so,

$x = y=80^o$

Hence,

$x = 80^{\circ}$

$y = 80^{\circ}$

$z = 80^{\circ}$

$t = 80^{\circ}$

1340192_1082942_ans_00302c79fcef49fc864b3b581e405e7f.png

One of the acute angles of a right triangle is $40°$ . Find the other acute angle.

In an right angled triangle,

${ \theta }_{ 1 }+{ \theta }_{ 2 }=90°$

$40°+{ \theta }_{ 2 }=90°$

${ \theta }_{ 2 }=50°$

Thus, the other acute angle is

$50°$

Two parallel lines $l$ and $m$ are cut by a transversal $t$ . If the interior angles of the same side of $t$ are $(2x-8)^\circ$ and $(3x-7)^\circ$ , find the measure of each of these angles.

Using the property of interior angles on same sides

$\begin{array}{l}2x - 8 + 3x - 7 = 180\\5x - 15 = 180\\x = {39^ \circ }\\2x - 8 = 2 \times 39 - 8 = 70\\3x - 7 = 3 \times 39 - 7 = 110\end{array}$

Find the complement of: $\dfrac{1}{2}$ of $40^0$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{1}{2}$ of $40^{o}$ is,

$=90^{o}-(\dfrac{1}{2} \times 40^{o})$

$=90^{o}-20^{o}$

$=70^{o}$ .

Therefore, the complement angle of $\dfrac{1}{2}$ of $40^{o}$ is $70^{o}$ .

An angle $10^{0}$ more than three times its complementary angle. Then find the angles.

Let one angle be

$x$ , then larger angle is

$3x+10^\circ$ .

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

$\implies$

$x+3x+10^\circ=90^\circ$

$\implies$

$4x=80^\circ$

$\implies$

$x=20^\circ.$

Hence, the smaller angle is

$x=20^\circ$

and larger angle is

$3x+10^o=3(20^o)+10^o=60^o+10^o=70^\circ.$

In $\Delta PQR$ S is a point on PQ such that SR = QR = SP. and $\angle PSR=120^{\circ}$ .Find $\angle P$ .

$\Delta PRS$

$PS=SR\implies \angle P=\angle R=x$

$\angle S=120$

By Angle sum property

$x+x+120=180$

$2x=180-120$

$x=\dfrac {60}2=30$

Prove that the line segments joining the points of contact of two parallel tangents of a circle,passes through its centre ?

Let,

$PQ$ and

$RS$ be two parallel tangents to a circle with cantre

$O$ where

$M,N$ be the points of contact of the two tangents

$PQ,RS$ to the circle respectively.

We have to prove that the line

$MN$ passes through the centre

$O$ of the circle.

Here,

$OM$ and

$OA$ are joined.

$OA\parallel PQ\\ PM\parallel AO\\ \therefore \angle PMO+\angle AOM={ 180 }^{ o }\quad \quad \left[ sum\quad of\quad two\quad adjacent\quad interior\quad angles\quad \right]$

but,

$\angle PMO=90^{ o }\quad \quad \left[ \because A\quad tangent\quad to\quad a\quad circle\quad is\quad perpendicular\quad to\quad the\quad radius\quad through\quad point\quad of\quad contact \right]$

So,

$90^{ o }+\angle AOM={ 180 }^{ o }\\ \Rightarrow \angle AOM=90^{ o }\\ \therefore \angle AON=90^{ o }\\ \angle AOM+\angle AON={ 180 }^{ o }$

$\therefore$

$MN$ is a straight line and it passes through

$O$ which is the centre of the given circle.

Hence, proved.

1149407_1102860_ans_a58a2bb0c6af4d8495da2a65684da1e2.jpg

If
A,B,C are the interior angles of triangle ABC, then prove that $\text{cosec}\left( {\dfrac{{C + A}}{2}} \right) = \sec \left( {\dfrac{B}{2}} \right)$

$\because \angle A ,\angle B$ and

$\angle C$ are angles of a triangle

$\implies \angle A+\angle B+\angle C=180^{\circ}$ (angle sum property of a triangle)

$\implies \angle A+\angle C=180^{\circ}-\angle B$

$\implies \dfrac{\angle A+\angle C}{2}=90^{\circ}-\dfrac{\angle B}{2}$ (on dividing both sides by

$2$ )

$\implies \text{cosec}\left(\dfrac{\angle A+\angle C}{2}\right)=\text{cosec}\left(90^{\circ}-\dfrac{\angle B}{2}\right)$ (on applying

$\text{cosec}$ both the sides)

$\implies$

$\text{cosec}\left( {\dfrac{{\angle A + \angle C}}{2}} \right) = \sec \left( {\dfrac{\angle B}{2}} \right)$

$[\because \text{cosec}(90^{\circ}-\theta)=\sec \theta]$

Hence proved.

An angle is $14^{o}$ more than its complementary angle. What is the measure?

Let the first angle is $={{x}^{o}}$

Then, the second angle is $=x+{{14}^{o}}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

According to given question,

$x+x+14={{90}^{o}}$

$2x={{90}^{o}}-{{14}^{o}}$

$2x={{76}^{o}}$

$x={{38}^{o}}$ .

Hence, the angles are $x=38^o$

and $x+{{14}^{o}}={{38}^{o}}+{{14}^{o}}={{52}^{o}}$ .

In the given figure line AB and CD intersect at O . $\angle BOC = {36^ \circ }$ . Find $\angle x$ , $\angle y$ and $\angle z$

In first figure,

$\angle AOD=\angle BOC$ ------- Vertically Opposite angles

$\therefore x= 36^\circ$

$\angle AOC+\angle COB=180^\circ$ ------ Angles on straight line

$y+36^\circ=180^\circ$

$y=180^\circ-36^\circ$

$y=144^\circ$

$\angle AOC=\angle BOD$ ------- Vertically Opposite angles

$y=z=144^\circ$

In second figure,

$\angle POR+\angle ROT+\angle QOT=180^\circ$ ------ Angles on straight line

$2x+90^\circ+x=180^\circ$

$3x=90^\circ$

$x=30^\circ$

$\therefore 2x= 60^\circ$

$\angle POR=\angle SOQ$ ------- Vertically Opposite angles

$2x=y=60^\circ$

$\angle POS=\angle ROQ$ ------- Vertically Opposite angles

$z=90^\circ+x$

$z=90^\circ+30^\circ$

$z=120^\circ$

In the above sided figure,if $QT \bot PR.,\angle TQR=40^o$ and $\angle SPR=30^o$ ,find $x$ and $y$

$\Delta QTS$ , the nagles are:

$40^o, 90^o, x$

Using sum of angles of triangle property:

$40^o+90^o+x=180^o\implies=50^o$

$y$ is and exterior angle of

$\Delta PSR$ , using the property of exterior angle:

$y = 30^o + x = 80^o$

${x=50^o, y=80^o}$

In the joining figure, name the following pairs of angles.
Adjacent angles that do not form a linear pair.

Consider the given figure.

Name of the angles that do not form linear pair are:

(a) $\angle BOA$ and $\angle AOE$

(b) $\angle AOE$ and $\angle EOD$

Hence, this is the answer.

$\angle{A}$ and $\angle{B}$ are supplementary angles. $\angle{A}=x+{10}^{o}$ and $\angle{B}=x-{10}^{o}$ , find $\angle{A}$ and $\angle{B}$

$\angle A=x+10^{\circ},\angle B=x-10^{\circ}$

Given

$\angle A$ and

$\angle B$ are supplementary angles

$\implies x+10^{\circ}+x-10^{\circ}=180^{\circ}\implies x=90^{\circ}$

$\angle A=90^{\circ}+10^{\circ}=100^{\circ}$

$\angle B=90^{\circ}-10^{\circ}=80^{\circ}$

The measure of an angle is twice the measure of its supplementary angle. Find its measure.

Solution:

Let the angle be

$X$

the supplement

$={180}^{\circ}-x$

According to the question

$x=2({180}^{\circ}-x)$

$3x={360}^{\circ}$

$x={120}^{\circ}$

Hence the angle is

${120}^{\circ}$

Find the value of $x$ in each of the following figures ?

The interior angle of angle a =180°-110°=70°
the interior angle of angle b=180°-120°=60°
the sum of the angles of a triangle is 180°
70°+60°+x°=180°
130°+x°=180°
x°=50°
therefore x =50°.

Write the complement of ${ 51 }^{ \circ } ,28^{ \circ },13^{ \circ }$ ?

We know that, if two angles are complementary, then their sum is

$90^o$ .

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

The complement of

$x={ 51 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 51 }^{ o }={ 39 }^{ o }.$

The complement of

$x={ 28 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 28 }^{ o }={ 62 }^{ o }.$

The complement of

$x={ 13 }^{ o }$ is

$90^o-x={ 90 }^{ o }-{ 13 }^{ o }={ 77 }^{ o }.$

The two angles $\left(3x+10\right)^{o}$ and $\left(2x+30\right)^{o}$ will make a straight line then find $x$ ?

Two angles

$(3x+10)^{o}$ and

$(2x+30)^{o}$ make straight line

$\therefore 3x+10+2x+30= 180^{o}$

$\therefore 5x+40 = 180^{o}$

$\therefore 5x = 140$

$\therefore x = \dfrac{140}{5} =28^{o}$

$x= 28^{o}$

In the figure. $A B \parallel C D$ , $\angle P T B = 55 ^ { \circ }$ and $\angle D V S = 45 ^ { \circ }$ .
Find sum of $\angle \mathrm { CUQ }$ and $\angle R T P$ is :

Given

$AB\parallel CD\angle PTB=55^{o}, \angle DVS=45^{o}$

To find :

$\angle CUQ+\angle RTP$

$AB\parallel CD$ then,

$\angle VTB=\angle DVS=45^{o}$ [corresponding angles]

According to lines pair of angles

$\angle RTP+\angle PTB+\angle VTB=180^{o}$

$\angle RTP+55^{o}+45^{o}=180^{o}$

$\angle RTP=80^{o}$

Then,

$\angle ATU=\angle PTB=55^{o}$ (vertically opposite angles)

$AB\parallel CD$ , then

$\angle CUQ=\angle ATC=55^{o}$

$\therefore \angle CUQ +\angle RTP=55^{o}+80^{o}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =135^{o}$

1363526_1179034_ans_58f0c8db31e64e3ea1fdf8ae9e306c2e.png

Fill in the blanks
An angle which measure is $180^{o}$ is called ..

An angle which measures is

$180^{o}$

is called Straight angle

In the given figure $\Box PTQR$ and PS is the bisector of $\angle P.$ If $\angle Q = 60^o$ and $\angle R = 30^o$ , find $\angle TPS$ .

$\triangle PQR, \angle P=180-(\angle Q+\angle R)=180-(60+30)=90^{\circ}$

$\triangle QPT, \angle P=180-(\angle Q+\angle T)=180-(60+90)=30^{\circ}$

$\triangle PTS, \angle S=180-(\angle T+\angle P)=180-(90+30)=60^{\circ}$

$\therefore \angle TPS=30^{\circ}$

Three angles which add upto $180^o$ are in the ratio $2:3:7$ . Find them.

Let the 3 angles be

$2x,3x,7x$

So,

$2x+3x+7x=180^o\Rightarrow 12x=180^o\Rightarrow x=15^o$

So, Angles will be

$2x=2*15=30^o\\3x=3*15=45^o\\7x=7*15=105$

Prove that the sum of angles of a triangle is 180.

$\textbf{Step -1: Prove that the sum of angles of a triangle is }\mathbf{180^\circ.}$

$\text{Let a triangle }ABC\text{ and draw a line }l\text{ through }A\text{ and parallel to the base }BC.$

$\because BC\parallel l\text{ and }AB \text{ and }AC\text{ are transversals on it.}$

$\therefore \angle 4=\angle 2\ldots(i)$ $\textbf{(Alternate angles)}$

$\text{and } \angle 5=\angle 3\ldots(ii)$ $\textbf{(Alternate angles)}$

$\text{Adding equations }(i)\text{ and }(ii),\text{ we get}$

$\angle 4+\angle 5=\angle 2+\angle 3$

$\text{Add }\angle 1\text{ on both sides.}$

$\Rightarrow \angle 1+\angle 4+\angle 5=\angle 1+\angle 2+\angle 3\ldots(iii)$

$\angle 1+\angle 4+\angle 5=180^\circ$ $\mathbf{(\because\textbf{l is a straight line})}$

$\text{From }(iii),$

$\angle 1+\angle 2+\angle 3=180^\circ$

$\text{Hence, sum of angles of a triangle is }180^\circ.$

$\textbf{ Hence Proved.}$

In fig, sides $QP$ and $RQ$ of triangle $PQR$ are produced to point $S$ and $T$ respectively. If angle $S P R = 135 ^ { \circ }$ and angle $P Q T = 110 ^ { \circ } \text { ,find } \angle P R Q$ .

REF.Image

So:-

$\angle PQT+\angle PQR= 180$

$110+ \angle PQR= 180$

$\angle PQR = 70^{\circ}$

Now,

$\angle SPR+\angle QPR= 180$

$135^{\circ}+\angle QPR= 180$

$\angle QPR= 45^{\circ}$

Now,

$\angle QPR+\angle PQR+\angle QRP= 180$

$45+70+\angle QRP = 180$

$\angle QRP = 65$

1104945_1181761_ans_bf7c1890558e46bf9a829c8c01e0c84a.jpg

Two supplementary angles differ by $22 ^ { 0 }$ . Find the angles.

Let first angle be

$x^{o}$ .

And, Then, second angle =

$180^{o}-x^{o}$

According to question,

$x^{o}-(180^{o}-x^{o})=22^{o}$

$2x^{o}=202^{o}$

$x=101^{o}$

Therefore, the required angles are

$101^{o}$ and

$79^{o}$ .

$OC$ and $OD$ are two rays from point $O$ on straight line $AOB$ . $\angle AOC =2x, \angle COD = 3x, \angle BOC=5x$ .
Find each angle : $\angle AOC , \angle COD, \angle DOB$

$\angle BOC=5x=\angle COD+\angle DOB$

$\Rightarrow 5x=3x+\angle DOB$

$\Rightarrow \angle DOB=2x$

Now

$\angle AOC+\angle COD+\angle DOB=180^o$

$\because AOB$ is

$\Rightarrow 2x+3x+2x=180^o$

$\Rightarrow 7x=180^o$

$\Rightarrow x=\left(\dfrac{180}{7}\right)^o$

$\Rightarrow \angle AOC=2x=\left(\dfrac{360}{7}\right)^o$

$\angle COD=3x=\left(\dfrac{540}{7}\right)^o$

$\angle DOB=2x=\left(\dfrac{360}{7}\right)^o$

1349871_1222288_ans_f0a2f9059e4e4ee392b90d81c4b14593.png

Find the measure of the angles of a triangle in each of the following cases:
One of the acute angles of a right triangle is $63^{\circ}$ . Find the other angle.

Right triangle has one angle equals to

$90^\circ$

So,

$90^\circ$

$63^\circ$

According to angle sum property of a traingle,

$\begin{array}{c}A + B + C = 180^\circ \\90^\circ + 63^\circ + C = 180^\circ \\C = 27^\circ \end{array}$

Thus Other angle is

$27^\circ$

Find all the angle of triangle ABC

$\Delta ABC$ , we have

$AB=AC$

$\Rightarrow \angle C=\angle B$ ........(1)

[

$\because$ Angles opposite to equal sides of a triangle are equal]

$BC=AC$

$\Rightarrow \angle A=\angle B$ ........(2)

[

$\because$ Angles opposite to equal sides of a triangle are equal]

Now,

$\angle A+\angle B+\angle C=180^0$ [

$\because$ Angle sum property of a triangle ]

$\Rightarrow \angle A+\angle A+\angle A=180^0$ [

$\because$ From (1) and (2) ]

$\Rightarrow 3\angle A=180^0$

$\Rightarrow \angle A=\dfrac{180^0}{3}=60^0$

$\therefore \angle A+\angle B+\angle C=60^0$

Find $\angle x = ?$

Sum of angles in triangle

$=180^{o}$

$\therefore \angle A+ \angle B+ \angle C=180^{o}$

$\Rightarrow \angle A+60+60=180^{o}$

$\Rightarrow \angle A=60^{o}$

Now

$x+ \angle A =180$ [Linear Angle]

$\therefore x=180-60=120^{o}$

1350194_1218741_ans_b15741af5a014b10bbadef9460776348.png

Find the angles in each of the following:
The angles are complementary and the smaller is $40 ^ { \circ }$ less than the larger.

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

Then, if one angle is

$x$ , then its complement given is

$x+40^\circ$ .

Then,

$x+x+40^\circ=90^\circ$

$\implies$

$2x=90^o-40^o=50^o$

$\implies$

$x=\dfrac{90^o}{5}=18^o$ .

Then, smaller angle

$x=25^\circ$ and larger angle is

$x+40^o=25^o+40^o=65^\circ.$

In the given figure , $AOC$ is the diameter of the circle , with centre $O$ .
If arc $AXB$ is half of arc $BYC$ , find $\angle BOC$ .

1117761_1208260_ans_b97a03d888f5413aabdb20a963e51b1e.jpg

Observe the figure and find the value of $x$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then,

$\angle COB+\angle COA=90^{\circ}$

$\Rightarrow 3x+2x=90^{\circ}$

$\Rightarrow 5x=90^{\circ}$

$\Rightarrow x=18^{\circ}$ .

Find the angle which is supplementary of itself.

Two angles are supplementary when sum of them is

$180^0$

Let

$x$ be the angle.

If supplementary of

$x$ is

$x$

$x+x=180\Rightarrow 2x=180$

$x=90^0$ is angle which is

supplementary of itself.

The measure of an angle which is $9$ times its supplement is?

$\\Let\>angle\>be\>x\>then\>its\>supplementary\>angle\\=(180^0-x)\\now\\x=9(180^{\circ}-x)\\x=9\times\>180^{\circ}-9x\\\therefore\>10x=9\times180^{\circ}\\\therefore\>x=(\dfrac{9\times180^{\circ}}{10})=162^{\circ}$

Find the value of $x$

As it is a straight line -

$\Rightarrow 3x+6x=180°$

$9x=180°$

$x=20°$

Hence, the answer is

$20°$

1188731_1223145_ans_c3bfa2327e02438ab3ca610cbf0a7f7b.png

Angles Q and R of a $\Delta PQR$ are $25^{\circ}$ and $65^{\circ}$ . Write which of the following is true:
1. $PQ^2 +QR^2 = RP^2$ 2. $PQ^2 +RP^2 = QR^2$ 3. $RP^2 +QR^2 = PQ^2$

Given,

$\angle Q=25^o$

$\angle R=65^o$

Using the angle sum property, i.e.,

$\angle P+\angle Q+\angle R=180^o$

$\Rightarrow \angle P=180^o-\angle Q-\angle R$

$=180^o-25^o-65^o$

$=180^o-90^o$

$=90^o$

$\therefore\Delta PQR$ is a right angled triangle.

$\therefore$ By Pythagoras theorem,

$QR^2=PQ^2+RP^2$

Hence option

$2$ , is correct.

1368828_1235963_ans_635cf8e7584144f88030e61312d37e5b.JPG

The measures of two complementary angles are $x$ and $4x$ . Find the measure of each of these angles.

We know, the sum of two complementary angles is

$90^o$ .

Given, the complementary angles are

$x$ and

$4x$ .

Then,

$x+4x=90^o$

$\implies$

$5x=90^o$

$\implies$

$x=18^o$ .

Therefore, the angles are

$x=18^o$ and

$4x=4\times18^o=72^o$ .

In the adjoining figure, $ABCD$ is a rectangle whose diagonals $AC$ and $BD$ intersect at $O$ . If $\angle AOB=110^{o}$ , find the measure of: $\angle AOD$ and $\angle BOC$ .

Apply the concept of supplementary angles.

$\angle AOB+\angle AOD=180^\circ$

$110^\circ+\angle AOD=180^\circ$

$\angle AOD=70^o$

$\angle AOD=\angle BOC$ vertically opposite angles.

Thus,

$\angle BOC=70^\circ$

An angle is $\dfrac{2}{3}$ of its complement. What is its magnitude?

We have, angle is

$\dfrac {2}{3}$ of its complement.

Let the required angle be

$x$ .

Then, its complement will be

$=90^o-x$ .

Given, that

$x=\dfrac {2}{3}(90^o-x)$

$\implies$

$3x=2\times (90^o-x)$

$\implies$

$3x=180^o-2x$

$\implies$

$3x+2x=180^o$

$\implies$

$5x=180^o$

$\implies$

$x=\dfrac {180^o}{5}$

$\implies$

$x=36^o$ .

$\therefore$ Its complement

$=90^o-x$

$=90^o-36^o$

$=54^o$ .

Hence, the required angle is

$36^o$ .

In the adjoining Figure, $\overline{XY}\parallel \overline {BC}$ . Find the values of $x, y, z$ .

Given that XY

$||$ BC

then,

$\Delta ABC$

$\angle A+\angle B+\angle C=180^o$ (property of triangle)

$40^o+30^o+\angle C=180^o$

$70^o+z=180^o$

$z=180^o-70^o$

$z=110^o$

$XY||BC$

then

$\angle x^o=\angle B$

$\angle x^o=30^o$ .

Now, In

$\Delta AXY$

$\angle x^o+\angle y^o+\angle A=180^o$ (property of triangle)

$30^o+\angle y^o=40^o=180^o$

$\angle y^o=180^o-70^o$

$\angle y^o=110^o$

Hence, this is the answer.

1195693_1241922_ans_21a13a304f4e4896ba9f85d2593f7909.png

Two angles are equal and they form a pair of supplementary angles. Find their magnitude.

We have,

$M\angle DBA =40^o$

$M\angle DBC =50^o$

$M\angle ABC =90^o$

$\therefore \ \angle DBA$ and

$\angle ABC$ are supplementary angles.

$\therefore \ M\angle DBA +M\angle DBC=90^o$

Hence, this is the answer.

1345486_1240689_ans_fd47d2aef3e34e6c8d02856673e4e4ea.png

Find the value of $x$ in each of the following figure :

As we know that

The sum of all angle of triangle is

$180^0$

Now,

$\angle A=35^0$

$\angle B=\angle D$ alternate angle

$35^0+48^0+x=180^0$

$83^0+x=180^0$

$x=180^0-83^0$

Hence,

$x=97^0$

In fig name all the pairs of adjacent angles.

$\begin{array}{l} Adjacent\, \, park \\ \angle ABE\, \, \& \, \, \angle ABC \\ \angle ACb\, \, \& \, \, \, \angle BCF \\ \angle BAC\, \, \& \, \, \angle DAC \\ \angle ADB=\angle ADC \end{array}$

Find the values of $x$ and $y$ in the given figure

Sum of angles of a

$\Delta =180^o$

For

$\Delta ADB$

$90^o+45^o+y=180^o$

$y=180-135$

$=45^o$ .

For

$\Delta ADC$

$x+40^o+90^o=180^o$

$x=180-90-40$

$=90-40$

$=50^o$ .

Find the value of x

REF.Image

We have,

$3x+4x=90^{\circ}$

$\Rightarrow 7x=90^{\circ}$

$x=\frac{90^{\circ}}{7}$

$\Rightarrow x=12.85$ (approx)

1211614_1306240_ans_dc7e11a7a9d64437bea4c14861fff331.JPG

Find the angle which is $\dfrac{1}{5}$ of its complement.

We have, angle is

$\dfrac {1}{5}$ of its complement.

Let the one angle be

$x$ .

Then, its complement will be

$=90^o-x$ .

Given, that

$x=\dfrac {1}{5}(90^o-x)$

$\implies$

$5x=90^o-x$

$\implies$

$5x=90^o-x$

$\implies$

$5x+x=90^o$

$\implies$

$6x=90^o$

$\implies$

$x=\dfrac {90^o}{6}$

$\implies$

$x=15^o$ .

$\therefore$ Its complement

$=90^o-x$

$=90^o-15^o$

$=75^o$ .

Hence, the angle is

$15^o$ .

If the complement of an angle is $27^{\circ}$ , then find its supplement.

Complement of an angle

$= 27^o$

Let the angle be

$x^o$

then,

$x^o + 27^o = 90^o$

$\therefore x^o = 90^o - 27^o$

$\therefore x = 63^o$

Now, supplement of the angle

$= 180^o - x^o = 180^o - 63^o$

$= 117^o$

Two complementary angles are in the ratio $2:3.$ Find the supplement of a larger angle.

1371910_1266675_ans_c31c09b6033c457ca5913c1c21547aba.PNG

In Fig., $\Delta ODC \sim \Delta OBA,\angle BOC = {125^ \circ }$ and $\angle CDO = {70^ \circ }$ . Find $\angle DOC,\angle DCO$ and $\angle OAB$

By linear angle property,

$\angle COD+\angle COB=180^o$

$\Rightarrow \angle COD=180-125=55^o$

$\Delta COD, 55+70+\angle DCO=180^o$

$\Rightarrow \angle DCO=55^o$

$\Delta ODC\sim \Delta OBA$

$\Rightarrow \angle OCD=\angle OAB=55^o$ .

1194168_1265901_ans_2dba6bafb6d94341a9f66a2f0d1546a4.jpg

Write down each pair of adjacent angles shown in fig.

$\begin{array}{l} Adjacent\, \, angle \\ \angle DOC\, \, \& \angle COB \\ \angle COB\, \, \& \angle BOA \\ \angle DOB\, \, \& \angle BOA \\ \angle DOC\, \, \& \angle COA \end{array}$
1204444_1296254_ans_8190f12474bc452786a567e7a4e18cfa.PNG

1204444_1296254_ans_8190f12474bc452786a567e7a4e18cfa.PNG

Find the complementary angle of $45\dfrac{1}{2}^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $37\dfrac{1}{2}^o=\dfrac{91}{2}^o$ .

Then, the complement of $\dfrac{91}{2}^{0}=90^{0}-\dfrac{91}{2}^{0}=\dfrac{89}{2}^{0}=44\dfrac{1}{2}^0$ .

In the adjoining figure, identify the pair of corresponding angles.

The pairs of corresponding angles:

$\angle 1, \angle 5; \, \angle 2, \angle 6; \, \angle 4 \angle 8$ and

$\angle 3, \angle 7$

In the figure, if $a-b=80^{o}$ , find $a$ and $b$ .

$a - b = 80° ..... \left( 1 \right) \quad \left( \text{Given} \right)$

$a + b = 180° ..... \left( 2 \right) \quad \left( \text{Linear pair of angles} \right)$

Adding equation

$\left( 1 \right) \& \left( 2 \right)$ , we have

$2a = 260°$

$\Rightarrow a = \cfrac{260°}{2} = 130°$

Substituting the value of

$a$ in equation

$\left( 1 \right)$ , we have

$130° - b = 80°$

$\Rightarrow b = 130° - 80° = 50°$

In the adjoining figure, identify the pair of alternate interior angles.

The pairs of alternate interior angles are:

$\angle 3, \angle 5$ and

$\angle 2, \angle 8$

An angle is $28^{o}$ less than its complement, find its measure.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the angle be ${x}^{\circ}$ and its complement is ${90}^{\circ}-{x}^{\circ}$

Then, according to the statement,

$\Rightarrow {x}^{\circ}=\left({90}^{\circ}-{x}^{\circ}\right)-{28}^{\circ}$

$\Rightarrow {x}^{\circ}+{x}^{\circ}=90^o-{28}^{\circ}$

$\Rightarrow 2{x}^{\circ}={62}^{\circ}$

$\Rightarrow {x}^{\circ}=\dfrac{{62}^{\circ}}{2}={31}^{\circ}$ .

Hence, the measure of the angle is $x={31}^{\circ}$

and its complement is $90^o-x=90^o-31^o={59}^{\circ}$ .

Find the complementary angle of $37\dfrac{1}{2}^o.$

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $37\dfrac{1}{2}^o=\dfrac{75}{2}^o$ .

Then, the complement of $25^{0}=90^{0}-\dfrac{75}{2}^{0}=\dfrac{105}{2}^{0}=52\dfrac{1}{2}^0$ .

An angle is equal to $5$ times its complement. Find the measure of each angle.

It is given that the angle is $5$ times its complement.

Consider $x$ and $5 x$ as the angles.

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, $x +5x = 90^0$

$\implies$ $6x = 90^0$

$\implies$ $x = \dfrac{90}{6} = 15^0$ .

Here, the angles will be $x= 15^0$ and $5x=5 \times 15^o = 75^0$ .

What is an Obtuse Angle ?

An angle which is lying between

$90^{o}$ and

$180^{o}$ is called Obtuse Angle.

One of the angle of a triangle is ${ 50 }^{ \circ }$ and the other two angles are equal. Find the measure of each of the equal angles.

Let the equal angles be

$x, x$

Then by angle property:

$\Rightarrow x+ x+50^o=180^o$

$\Rightarrow 2x=130^o$

$\Rightarrow x=\dfrac{130^o}{2}$

$\Rightarrow x=65^o$

Angles are

$65^o, 65^o$

If the angle $( 2 x - 10 ) ^ { \circ }$ and $( x - 5 ) ^ { \circ }$ are complementary angles. Find the angles.

It is given that two complementary angles are $(2x-10)^o$ and $(x-5)^o$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, $(2x-10)^o$ $+$ $(x-5)^o$ $= 90^0$

$\implies$ $2x-10^o+x-5^o = 90^0$

$\implies$ $3x-15^o = 90^0$

$\implies$ $3x=15^o +90^0=105^o$

$\implies$ $x = \dfrac{105}{3} = 35^0$ .

Here, the angles will be $(2x-10)^0=2(35^o)-10^o=70^o-10^o=60^o$

and $(x-5)^0=(35^o)-5^o=30^o$ .

Find the complementary angle of ${28}^{\circ}.$

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the measure of its complement be $x.$

Then,

$x + 28^\circ = 90^\circ$

$\Rightarrow x = 90^\circ - 28^\circ$

$\Rightarrow x = 62^\circ$

Hence, the complement of the given angle is $62^\circ.$

Show that the lines $2x+y+1=0,\,\,x-2y+1=0$ are perpendicular.

First line

$2x+y+1=0$ $

Slope

$\,{m}_{1}=\dfrac{-\,coefficient\,of\,x}{coefficient\,of\,y}=\dfrac{-2}{1}=-2$

Second line is

$x-2y+1=0$

Slope

$\,{m}_{2}=\dfrac{-\,coefficient\,of\,x}{coefficient\,of\,y}=\dfrac{-1}{-2}=\dfrac{1}{2}$

$\therefore \,{m}_{1}\times {m}_{2}=-2\times \dfrac{1}{2}=-1$

$\therefore$ Given lines are perpendicular.

Find the complement of: ${23}^{\circ}{19}^{\prime}{32}^{\prime\prime}$ .

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

Let the angle be ${x}^{\circ}$ and its complement is ${90}^{\circ}-{x}^{\circ}$

Then, the complement of $x={23}^{\circ}{19}^{\prime}{32}^{\prime\prime}$

$={90}^{\circ}-{x}^{\circ}$

$={90}^{\circ}-{23}^{\circ}{19}^{\prime}{32}^{\prime\prime}$

$={89}^{\circ}{59}^{\prime}{60}^{\prime\prime}-{23}^{\circ}{19}^{\prime}{32}^{\prime\prime}$

$={66}^{\circ}{40}^{\prime}{28}^{\prime\prime}$ .

Hence, the complement is ${66}^{\circ}{40}^{\prime}{28}^{\prime\prime}$ .

Find the complement of: ${58}^{\circ}{20}^{\prime}{25}^{\prime\prime}$ .

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

Let the angle be ${x}^{\circ}$ and its complement is ${90}^{\circ}-{x}^{\circ}$

We can write,

${90}^{\circ}={89}^{\circ}{59}^{\prime}{60}^{\prime\prime}$ .

Then, the complement of

$x={58}^{\circ}{20}^{\prime}{25}^{\prime\prime}$

$={90}^{\circ}-{x}^{\circ}$

$={90}^{\circ}-{58}^{\circ}{20}^{\prime}{25}^{\prime\prime}$

$={89}^{\circ}{59}^{\prime}{60}^{\prime\prime}-{58}^{\circ}{20}^{\prime}{25}^{\prime\prime}$

$={31}^{\circ}{39}^{\prime}{35}^{\prime\prime}$ .

Hence, the complement is ${31}^{\circ}{39}^{\prime}{35}^{\prime\prime}$ .

In Fig. 7.12, if AB ||DC and E is the mid - point of side AC, then prove that E is the mid-point of side BD.

In the given figure,

$AB\parallel DC$

Hence,

$\angle CDB=\angle DBA$ (Alternate angles)

Now in triangles DEC and BEA

$\angle DEC=\angle BEA$ (Vertically opposite angles)

$\angle CDB=\angle DBA$ (Proved above)

$EC=EA$ (E is midpoint of AC)

Hence, by AAS criterion triangle DEC

$\cong$ triangle BEA

Then by c.p.c.t.

$DE=BE$

Therefore E is the midpoint of BD

The difference between the angle and its complement is ${36}^{\circ}$ . Find the angles.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the angle be

${x}^{\circ}$ and its complement is

${90}^{\circ}-{x}^{\circ}$

Then, according to the statement,

$\Rightarrow {x}^{\circ}-\left({90}^{\circ}-{x}^{\circ}\right)={36}^{\circ}$

$\Rightarrow {x}^{\circ}-{90}^{\circ}+{x}^{\circ}={36}^{\circ}$

$\Rightarrow 2{x}^{\circ}={36}^{\circ}+{90}^{\circ}={126}^{\circ}$

$\Rightarrow {x}^{\circ}=\dfrac{{126}^{\circ}}{2}={63}^{\circ}$ .

Hence, the measure of the angle is

${63}^{\circ}$ and its complement is

${27}^{\circ}$ .

The difference in the measures of the two complementary angles is $12^o$ . Find the measure of the angles.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the angle be ${x}^{\circ}$ and its complement is ${90}^{\circ}-{x}^{\circ}$

Then, according to the statement,

$\Rightarrow {x}^{\circ}-\left({90}^{\circ}-{x}^{\circ}\right)={12}^{\circ}$

$\Rightarrow {x}^{\circ}-{90}^{\circ}+{x}^{\circ}={12}^{\circ}$

$\Rightarrow 2{x}^{\circ}={12}^{\circ}+{90}^{\circ}={102}^{\circ}$

$\Rightarrow {x}^{\circ}=\dfrac{{102}^{\circ}}{2}={51}^{\circ}$ .

Hence, the measure of the angle is $x={51}^{\circ}$

and its complement is $90^o-x=90^o-51^o={39}^{\circ}$ .

Find the measure of an angle which is ${18}^{\circ}{2}^{\prime}{10}^{\prime\prime}$ less than its complement.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the measure of an angle be

${x}^{\circ}$ .

$\therefore$ The measure of its complement

$={90}^{\circ}-{x}^{\circ}$

We can write,

${90}^{\circ}={89}^{\circ}{59}^{\prime}{60}^{\prime\prime}$ .

Then,

$\left({90}^{\circ}-{x}^{\circ}\right)-{x}^{\circ}={18}^{\circ}{2}^{\prime}{10}^{\prime\prime}$

$\Rightarrow 2{x}^{\circ}={90}^{\circ}-{18}^{\circ}{2}^{\prime}{10}^{\prime\prime}$

$\Rightarrow 2{x}^{\circ}={89}^{\circ}{59}^{\prime}{60}^{\prime\prime}-{18}^{\circ}{2}^{\prime}{10}^{\prime\prime}={71}^{\circ}{57}^{\prime}{50}^{\prime\prime}$

$\Rightarrow {x}^{\circ}=\dfrac{{71}^{\circ}{57}^{\prime}{50}^{\prime\prime}}{2}={35}^{\circ}{58}^{\prime}{55}^{\prime\prime}$ .

The measure of the angle is

${35}^{\circ}{58}^{\prime}{55}^{\prime\prime}$ .

What are complementary and supplementary angles ?

Complementary angles :

Two angles whose sum is 90

$^{o}$ are called complementary angles.

Supplementary angles :

Two angles whose sum is 180

$^{o}$ are called supplementary angles.

In the figure if Line l || m and the line n is the transversal then find the value of a and b

Given:

$l\parallel m$

sum of the interior angles on the same side of transversal is

${180}^{\circ}$

$b+{52}^{\circ}={180}^{\circ}$

$\Rightarrow b={180}^{\circ}-{52}^{\circ}={128}^{\circ}$

And

$a={52}^{\circ}$

$\therefore a={52}^{\circ},\,\,b={128}^{\circ}$

Find the complementary angle of $22\dfrac {1}{2}^{o}$ .

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $22 \dfrac{1}{2}^\circ=\dfrac{45}{2}$ .

Let the measure of its complement be $x^\circ.$

Then,

$\implies$ $x + \dfrac{45}{2}^o = 90^o$

$\implies x = 90^o - \dfrac{45}{2}^o$

$\implies x = \dfrac{135}{2}^o=67\dfrac{1}{2}^\circ$

Hence, the complement of the given angle measures $67\dfrac{1}{2}^\circ$

In the given figure, line l || line m and line n || p . Find $\angle a,\angle b,\angle c.$ from the given measure of an angle .

Let us mark the points R and S on line n, T and U on line p, A and B on line l and C and D on line m.

Suppose the lines n and p intersect the line l at K and L respectively and line m at M and N respectively.

Since, the line l and line p intersect at L, then

$\angle KLN = \angle BLT$ (Vertically opposite angles)

$\Rightarrow \angle KLN = 45^\circ$

Since, line l

$\parallel$ line m and line p is a transversal intersecting them at L and N, then

$\angle KLN + \angle MNL = 180^\circ$ (Pair of interior angles on the same side of transversal is supplementary)

$\Rightarrow 45^\circ + \angle a = 180^\circ$

$\Rightarrow \angle a = 180^\circ - 45^\circ = 135^\circ$

Since, the line m and line p intersect at N, then

$\angle DNU = \angle MNL$ (Vertically opposite angles)

$\Rightarrow \angle b = \angle a = 135^\circ$

Since, line n

$\parallel$ line p and line m is a transversal intersecting them at M and N, then

$\angle NMS = \angle DNU$ (Corresponding angles)

$\Rightarrow \angle c = \angle b = 135^\circ$

1805546_1412553_ans_3d1aa53203ba43efb8dd4d2c01033004.png

Which of the pair given is complementary and which are not:
a) ${ 15 }^{ \circ },{ 75 }^{ \circ }$
b) $76^{ \circ },{ 47 }^{ \circ }$
c) $64^{ \circ },{ 26 }^{ \circ }$

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are:

a) $15^o$ and $75^o$ .

Then, their sum $=15^o+75^o=90^o$ .

Therefore, they are complementary.

b) $76^o$ and $47^o$ .

Then, their sum $=76^o+47^o=123^o$ .

Therefore, they are not complementary.

c) $64^o$ and $26^o$ .

Then, their sum $=64^o+26^o=90^o$ .

Therefore, they are complementary.

In fiq. AB divides $\angle DAC$ in the ratio 1:3 and AB $=$ DB. Determine the value of x.

REF.Image

AB divides

$\angle DAC$ in the ratio

$1:3$

and

$AB=DB$

$\angle EAC =108^{\circ}$

Let

$\angle DAB=x$ and

$\angle BAC=3x$ (Given)

$x+3x+\angle EAC=180^{\circ}$

$\Rightarrow 4x+108=180^{\circ}$

$\Rightarrow 4x=180^{\circ}-108^{\circ}$

$\Rightarrow 4x=72^{\circ}$

$\Rightarrow x=72/4$

$\therefore x=18^{\circ}$

1201527_1416349_ans_0b422d19e69f4df1badfd98cb2a5e179.png

Find the complementary of the angle $30^o$ .

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $30^\circ$ .

Let the measure of its complement be $x^\circ.$

Then,

$\implies$ $x + 30^o = 90^o$

$\implies x = 90^o - 30^o$

$\implies x = 60^\circ$

Hence, the complement of the given angle measures $60^\circ.$

The difference between the measures of the two angles of a complementary pair is ${ 40 }^{ \circ }$ . Find the measures of the two angles.

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

Let the angle be ${x}^{\circ}$ and its complement is ${90}^{\circ}-{x}^{\circ}$

Then, according to the statement,

$\Rightarrow {x}^{\circ}-\left({90}^{\circ}-{x}^{\circ}\right)={40}^{\circ}$

$\Rightarrow {x}^{\circ}-{90}^{\circ}+{x}^{\circ}={40}^{\circ}$

$\Rightarrow 2{x}^{\circ}={40}^{\circ}+{90}^{\circ}={130}^{\circ}$

$\Rightarrow {x}^{\circ}=\dfrac{{130}^{\circ}}{2}={65}^{\circ}$ .

Hence, the measure of the angle is $x={65}^{\circ}$

and its complement is $90^o-x=90^o-65^o={25}^{\circ}$ .

In the figure, $AB = AC$ and $OB$ and $OC$ are angle bisectors of $\angle B$ and $\angle C.$ Prove that $\angle BOC = \angle ACD$ .

Let

$\angle ABO=\angle OBC=x$ and

$\angle ACO=\angle OCB=y.$

As it is given that

$AB=AC.$ So,

$\angle B=\angle C$

$\Rightarrow 2x=2y$

$\Rightarrow x=y\quad\quad\quad\dots(i)$

$\Delta ABC$ ,

$\angle ABC+\angle BCA+\angle CAB=180^o$

$\Rightarrow 2x+2y+\angle CAB=180^o$

$\Rightarrow \angle A=180^o-2x-2y\quad\quad\quad\dots(ii)$

By using exterior angle property,

$\angle ACD = \angle A+\angle B$

$=180^o-2x-2y+2x$ [From

$(ii)$ ]

$=180^o-2y\quad\quad\quad\dots(iii)$

Now, by using angle sum property in

$\triangle OBC,$

$\angle OBC+\angle OCB+\angle BOC=180^o$

$\Rightarrow x+y+\angle BOC=180^o$

$\Rightarrow y+y+\angle BOC=180^o$ [From

$(i)$ ]

$\Rightarrow \angle BOC=180^o-2y\quad\quad\quad\dots(iv)$

From equations

$(iii)$ and

$(iv)$ we can conclude that,

$\angle BOC=\angle ACD$

Hence, proved.

For what value of $x+y$ in the given figure will ABC be a line? Justify you answer.

The value of

$x+y$ can't be find out with the given data.

No,

$ABC$ is not a line as it not straight, it is bent at point

$B$

Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle. Justify your answer.

No, each of these angles will be a right angle only they form a linear pair, ie., when the non-common arms of the given two adjecent angles are two opposites rays and common arms is perpendicular to them.

Find the complementary of $75^{\circ}$ .

We know, two angles are said to be complementary, if the sum of their measures is $90^\circ.$

The given angle is $75^\circ.$

Let the measure of its complement be $x.$

Then,

$x + 75^\circ = 90^\circ$

$\Rightarrow x = 90^\circ- 75^\circ$

$\Rightarrow x = 15^\circ$

Hence, the complement of the given angle is $15^\circ.$

One angle of a linear pair is twice the other. How much is each?

If two angles form a linear pair, then the sum of the degree measures of the angle is

$180^{\circ}$ .

Let

$x=$ measure of the smaller angle

Then

$2x=$ measure of the other angle.

So,

$x+2x=180^o$

$3x=180^o$

$x=\dfrac{180}{3}$

$\therefore x=60^{\circ}$

The measures of angles are

$60^{\circ}$ and

$120^{\circ}$ .

Two angles measure $\left( 55 ^ { \circ } + 3 a \right)$ and $\left( 115 ^ { \circ } - 2 a \right) \cdot$ If each is supplement of the other, then calculate the value of $a .$

Given

$(55^{\circ}+3 a),(115^{\circ}-2 a)$ are supplementary angles

$\implies 55^{\circ}+3 a+115^{\circ}-2 a=180^{\circ}$

$\implies 170^{\circ}+a=180^{\circ}$

$\implies a=180^{\circ}-170^{\circ}$

$\implies a=10^{\circ}$

Define Adjacent angles, Linear pair of angles. What is the sum of a linear pair of angles

Adjacent angles

$\rightarrow$ Two angles are Adjacent when they have a common side and a common vertex (corner point) and don't overlap.

Linear pair of angles

$\rightarrow$ A linear pair of angles is formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines.

The sum of a linear pair of angles is

$180°$ .

In the figure, $PQR$ is a straight line. Find $x$ . Then complete the following :
(i) $\angle AQB=$
(ii) $\angle BQP=$
(iii) $\angle AQR=$

In the given fig.,

$\angle{PQA}, \angle{AQB}$ and

$\angle{BQR}$ forms a linear pair.

Therefore,

$\angle{PQA} + \angle{AQB} + \angle{BQR} = 180°$

$\left( x + 20° \right) + \left( 2x + 10° \right) + \left( x - 10° \right) = 180°$

$4x + 20° = 180°$

$\Rightarrow x = \cfrac{180° - 20°}{4} = 40°$

Therefore,

$\angle{AQB} = 2x + 10° = 2 \times 40° + 10° = 90°$

$\angle{BQP} = \left( x + 20° \right) + \left( 2x + 10° \right) = 3x + 30° = 3 \times 40° + 30° = 150°$

$\angle{AQR} = \left( 2x + 10° \right) + \left( x - 10° \right) = 3x = 3 \times 40° = 120°$

Write the complementary, supplementary angles of $75^\circ$ .

Complementary angles are those angles whose sum is

$90^o$ .

$=90^o-75^o=15^o$

Complementary angles of

$75^o=15^o$

Two angles are called supplementary angles, if their sum of two right angles

$=180^o$

$=180^o-75^o=105^o$

Supplementary angles of

$75^o=105^o$ .

In the given figure, AB is diameter of a circle with centre O and chord ED is parallel to AB and $\angle EAB={ 65 }^{ o }.$ Calculate (i) $\angle EAB$ (ii) $\angle BED$ (iii) $\angle BCD$

1220214_1447706_ans_c76a14aee14a46b1a4f2b1207ce01f8a.jpg

Write down the measures of
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each).

$45^{\circ}, 70^{\circ}$

$105^{\circ}, 132^{\circ}$ .

Sushma says, "Complementary angle of ${ 48 }^{ 0 }$ is ${ 42 }^{ 0 }$ ". Do you agree with her? Give reasons.

We know, two angles are said to be complementary, if the sum of their measures is

$90^\circ.$

Given, the angles are

$48^o$ and

$42^o$ .

Then, their sum $=48^o+42^o=90^o$ .

Therefore, they are complementary.

Hence, Sushma is right.

In the figure, show that $\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$ .

We know that the sum of all the angles in triangle

$ACE$ is

$180^{\circ}$ .

$\angle A + \angle C + \angle E = 180^{\circ} .. (1)$
We know that the sum of all the angles in triangle

$BDF$ is

$180^{\circ}$ .

$\angle B + \angle D + \angle F = 180^{\circ} .. (2)$
Now by adding both equations

$(1)$ and

$(2)$ we get

$\angle A + \angle C + \angle E + \angle B + \angle D + \angle F = 180^{\circ} + 180^{\circ}$
On further calculation

$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$
Therefore, it is proved that

$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^{\circ}$ .

In a right-angled triangle, one of the acute angles measures $53^{\circ}$ . Find the measure of each angle of the triangle.

Consider

$ABC$ as a right-angled triangle with

$\angle C = 90^{\circ}$

We know that the sum of all the angles in a triangle is

$180^{\circ}$ .

So we can write it as

$\angle A + \angle B + \angle C = 180^{\circ}$

We can write it as

$\angle A + \angle B =180^{\circ} - \angle C$

By substituting the values

$\angle A + \angle B = 180^{\circ} - 90^{\circ}$

By subtraction

$\angle A + \angle B = 90^{\circ}$

Let us consider

$\angle A = 53^{\circ}$

By substituting the value of

$\angle A$ in

$\angle A + \angle B = 90^{\circ}$

We get

$53^{\circ} + \angle B = 90^{\circ}$

On further calculation

$\angle B = 90^{\circ} - 53^{\circ}$

By subtraction

$\angle B = 37^{\circ}$

Therefore,

$\angle A = 53^{\circ}, \angle 37^{\circ}$ and

$\angle C = 90^{\circ}$ .

If $\left(2x-14\right)$ and $11\left( x+7 \right)$ are supplementary angles then find the angles.

$2x-14+11(x+7)=180^o$ (

$\because$ linear pair)

$2x-14+11x+77=180^o$

$13x=180-63=117$

$\therefore x=\dfrac{117}{13}=9^o$

$\therefore 2x-14=18-14=4^o$

$11(x+7)=11(9+7)=176^o$

The angles are

$4^o$ and

$176^o$ .

If $a=3b$ , find a and b. (in degree)

1321988_1586547_ans_c94c46e477a343ae9e3d9cb725c9af28.jpg

Find the value of x from the adjacent figure

${\textbf{Step -1: Calculating the value of x }}$

$\Rightarrow \angle {\text{AOD + }}\angle {\text{DOC + }}\angle {\text{COB = 18}}{{\text{0}}^ \circ }$

$\Rightarrow 2x + 3x + 5x = {180^ \circ }$

$\Rightarrow 10x = {180^ \circ }$

$\Rightarrow x = \dfrac{{180}}{{10}}$

$\Rightarrow x = {18^ \circ }$

${\textbf{ Hence, the value of }}\mathbf{x}{\text{ is }}\mathbf{18^ \circ. }$

Find the complement of each of the following angle: ${ 55 }^{ \circ }$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $55^o$ .

Then, the complement of $55^{0}=90^{0}-55^{0}=35^{0}$ .

Write the supplement of each of the following angles :
${ 54 }^{ \circ }$

Two Angles are Supplementary when they add up to

$180^o$ .

Let the required angle be

$x^o$ .

$\therefore$

$x+54^o=180^o$

$\Rightarrow$

$x=180^o-54^o$

$\Rightarrow$

$x=126^o$

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that $\Delta OCD$ is an isosecleles triangle.

1308210_1609029_ans_303912bd632e452c8dacfc5df6cc2d29.png

Find the value of the unknown exterior angle $x$ in the following diagram.

Sum of all angles of a

$\Delta le = 180^o$

$50^o + 50^o + \angle ACB = 180^o$

$\therefore \angle ACB = 80^o$

Now,

$x + \angle ACB = 180^o$

$\therefore x = 100^o$

$Or$

Since, Exterior angle

$=$ Sum of interior opposite angles, therefore

$x=50^o+50^o=110^o$ .

1339203_1645596_ans_00124dc60ae14abda34c7abc9d68beab.png

If two adjacent angles of a parallelogram are $(5x-5)^{\circ}$ and $(10x+35)^{\circ}$ , then find the ratio of these angles.

1870087_1593114_ans_7ccb423f6313461ab2408558c6fd2486.jpg

Find the value of the unknown exterior angle $x$ in the following diagrams.

${\textbf{Step -1: Find the interior opposite angles}}{\textbf{.}}$

${\text{Here, interior angles are angles which comes inside the triangle}}{\text{.}}$

${\text{They are}}$

$\angle A$

${\text{and}}$

$\angle B$

${\text{having}}$

$30^\circ$

${\text{and}}$

$60^\circ$

${\text{respectively, for named angles}}{\text{.}}$

${\text{Also, they are opposite to the exterior angle named}}$

$x.$

${\textbf{Step -2: Finding the exterior angle}}{\textbf{.}}$

${\text{Exterior angle = sum of interior opposite angle}}$

$x = 30^\circ + 60^\circ$

$x = 90^\circ$

${\textbf{Hence, the value of exterior angle}}$

${\mathbf{x}}$

${\textbf{is}}$

${\mathbf{90^\circ.}}$

List three situations where reflex angles may be seen.

Step 1: To find

(1) Three situations where reflex angles may be seen

Step 2: Three situations

Reflex angles are those angles whose measure is greater than

$180^o$ .

Examples where we can see reflex angle are shown in the figure:

$1$ . Pizza

$2$ . Electric pole

$3$ . Prism

2136863_1644450_ans_97a81d1ffed44efca01e6edd88b0be87.jpg

What value of x makes the lines $\bar{AB}$ and $\bar{CD}$ parallel to each other?

$\bar AB\parallel \bar CD$

$(2 x-7)^{\circ}$ and

$(3 x+13)^{\circ}$ are supplementary angles

$\implies (2 x-7)^{\circ}+(3 x+13)^{\circ}=180^{\circ}$

$\implies (2 x-7+3 x+13)^{\circ}=180^{\circ}$

$\implies (5 x+6)^{\circ}=180^{\circ}$

$\implies 5 x=180-6$

$\implies x=\dfrac{174}{5}=34.8$

What will be the complement of $59^o$ ?

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $59^o$ .

Then, the complement of $59^{0}=90^{0}-59^{0}=31^{0}$ .

Find the value of the unknown exterior angle $x$ in the following diagram.

Sum of all angles of a

$\Delta = 180^o$

$60^o + 60^o + \angle ABC = 180^o$

$\therefore \angle ABC = 60^o$

Now,

$x + \angle ABC = 180^o$

$\therefore x = 120^o$

$Or$

Since, Exterior angle

$=$ Sum of interior opposite angles, therefore

$x=60^o+60^o=120^o$ .

1339199_1645580_ans_3ddc23e4f9b14e69b6844546d582ae70.png

Find the value of the unknown interior angle $x$ in the following figure.

$\textbf{Step -1: Find }\mathbf{\angle ACB.}$

$\angle ACB=180^\circ-75^\circ$ $\textbf{(Linear Pair)}$

$=105^\circ$

$\textbf{Step -2: Find x.}$

$\because\text{Sum of angles in a triangle is }180^\circ.$

$\therefore \text{In }\triangle ABC,$

$35^\circ+x+\angle ACB=180^\circ$

$\Rightarrow 35^\circ+x+75^\circ=180^\circ$

$\Rightarrow x=180^\circ-35^\circ-75^\circ$

$\Rightarrow x=40^\circ$

$\textbf{Hence, the value of unknown }\mathbf{x\text{ is }40^\circ.}$

Find the value of unknown $x$ in the following diagram.

$\Delta XYZ$ ,

$\angle ZXY+\angle XYZ+\angle YZX=180^o$ [By angle sum property of a triangle]

$\Rightarrow 30^o+110^o+x=180^o$

$\Rightarrow x+140^o=180^o$

$\Rightarrow x=180^o-140^o=40^o$ .
1339228_1645647_ans_b90694a2d4f84faabfcdbb6f4bcd2e96.png

Find the value of unknown $x$ in the following diagram.

Sum of interior angles of a

$\Delta$ is

$180^o$

$\angle P + \angle Q + \angle R = 180^o$

$50^o + x + x = 180^o$

$2x = 180^o - 50^o$

$2x = 130^o$

$\therefore x = 65^o$

1339382_1645648_ans_3cae180cdede400fa6cae28ac6c62196.png

Find the value of the unknown interior angle $x$ in the following figure.

$\textbf{Step -1: Find the value of x}$

$\text{Here in }\triangle ABC,$

$\text{Exterior angle = 80}^\circ$

$\text{Opposite interior angles are = 30}^\circ \text{ and }x^\circ$

$\text{We know that an exterior angle of a triangle is equal to the sum of the opposite interior angles }$

$\Rightarrow 80^\circ = 30^\circ + x^\circ$

$\Rightarrow x^\circ = 80^\circ - 30^\circ$

$\Rightarrow x^\circ = 50^\circ$

$\boldsymbol{\textbf{Hence, the value of }\mathbf{x^\circ = 50^\circ}}$

Find the values of the unknown $x$ and $y$ in the following diagram.

$\textbf{Step 1: Find the required angles applying suitable properties.}$

$\text{In the diagram, angles measuring y and}$

$80^\circ$

$\text{are vertically opposite angles.}$

$\text{We know, vertically opposite angles are equal .}$

$\therefore y=80^\circ......(1)$

$\text{We know, sum of interior angles of a triangles is }180^\circ.$

$\text{So, from the given triangle, }$

$\therefore x+y+50^\circ=180^\circ$

$\Rightarrow x+80^\circ=130^\circ$

$[\textbf{From (1)}]$

$\Rightarrow x=50^\circ$

$\mathbf{\textbf{Hence, the values of x and y are } 50^\circ\ and \ 80^\circ\ respectively.}$

Find the value of unknown $x$ in the following diagram.

$\Delta PQR$ ,

$\angle RPQ+\angle PQR+\angle RPQ=180^{\circ}$ [By angle sum property of a triangle]

$90^{\circ}+30^{\circ}+x=180^{\circ}$

$x+120^{\circ}=180^{\circ}$

$\Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ}$ .
1339226_1645646_ans_d4aea71f67744b899f0f3ace501b45c8.png

Find the value of the unknown interior angle $x$ in the following figure.

$\textbf{Step -1: Find the value of x}$

$\text{Here in }\triangle ABC,$

$\text{Exterior angle = 115}^\circ$

$\text{Opposite interior angles are = 50}^\circ \text{ and }x^\circ$

$\text{We know that an exterior angle of a triangle is equal to the sum of the opposite interior angles }$

$\Rightarrow 115^\circ = 50^\circ + x^\circ$

$\Rightarrow x^\circ = 115^\circ - 50^\circ$

$\Rightarrow x^\circ = 65^\circ$

$\boldsymbol{\textbf{Hence, the value of unknown interior angle }\mathbf{x^\circ = 65^\circ}}$

Find the value of unknown $x$ in the following diagram.

Sun of interior angles of a

$\Delta$ is

$180^o$

$\angle P + \angle Q + \angle R = 180^o$

$x + x+ x = 180^o$

$3x = 180^o$

$\therefore x = 60^o$

1339383_1645649_ans_7d1911f402c542b1876bc254b42b72c5.png

Find the value of unknown $x$ in the following diagram.

Sum of interior angle of a

$\Delta le$ is

$180^o$

$\angle P + \angle Q + \angle R = 180^o$

$2x + x + 90^o = 180^o$

$3x = 90^o$

$\therefore x = 30^o$

1339384_1645650_ans_9b836004fcaa4278a88e5b20bab85628.png

Find the values of the unknown $x$ and $y$ in the following diagram.

$\textbf{Step 1: Find the required angles applying suitable properties.}$

$\text{In the diagram angles measuring y and } 120^\circ \text{ are linear pair.}$

$\text{We know, linear pair are supplementary.}$

$\therefore y+120^\circ=180^\circ$

$\Rightarrow y=60^\circ......(1)$

$\text{We know, sum of interior angles of a triangles is }180^\circ.$

$\text{So, from the given triangle, }$

$\therefore x+y+50^\circ=180^\circ$

$\Rightarrow x+60^\circ=130^\circ$

$[\textbf{From (1)}]$

$\Rightarrow x=70^\circ$

$\mathbf{\textbf{Hence, the values of x and y are } 70^\circ\ and \ 60^\circ\ respectively.}$

Find the angle which is equal to its supplement.

Let

$x$ be two equal angles of its supplement.
Therefore,

$x + x =180^o$ [Supplementary angles]

$\Rightarrow 2x=180^0$

$\Rightarrow x=\dfrac {180^o}{2}= 90^o$
Thus,

$90^o$ is equal to its supplement.

Find the values of the unknown $x$ and $y$ in the following diagram.

$\textbf{Step 1: Find the required angles applying suitable properties.}$

$\text{In the diagram, angles measuring x and}$

$60^\circ$

$\text{are vertically opposite angles.}$

$\text{We know, vertically opposite angles are equal .}$

$\therefore x=60^\circ......(1)$

$\text{We know, sum of interior angles of a triangles is }180^\circ.$

$\text{So, from the given triangle, }$

$\therefore x+y+30^\circ=180^\circ$

$\Rightarrow y+60^\circ=150^\circ$

$[\textbf{From (1)}]$

$\Rightarrow y=90^\circ$

$\mathbf{\textbf{Hence, the values of x and y are } 60^\circ\ and \ 90^\circ\ respectively.}$

In the given figure, $\angle 1$ and $\angle 2$ are supplementary angles. If $\angle 1$ is decreased, what changes should take place in $\angle 2$ so that both the angles still remain supplementary?

${\textbf{Step - 1: Defining supplementary angles}}$

${\text{Two angles are said to be supplementary if their sum is equal to 180}}^\circ {\text{.}}$

${\text{For example, x and y are two angles, then they are supplementary if x + y = 180}}^\circ$

${\textbf{Step - 2: Analyzing and answering the question}}$

${\text{The given angles are supplementary, i}}{\text{.e}}{\text{., }}\angle {\text{1 + }}\angle {\text{2 = 180}}^\circ$

$\angle {\text{1 - 1 + }}\angle {\text{2 + 1 = 180}}^\circ$ $\quad \quad \textbf{[Add and subtract by 1 on both sides]}$

${\text{This implies that if }}\angle {\text{1 is decreased and both are still supposed to remain supplementary, }}\angle {\text{2 must}}$

${\text{ increase by the same amount }}\angle {\text{1 is decreased}}{\text{.}}$

${\textbf{Hence, if angle 1 is decreased, angle 2 must be increased so that }}$ ${\textbf{they remain supplementary}}{\text{.}}$

Find the supplement of each of the following angles:

Supplementary angle of

$x = 180 - x$

Now, supplementry angle of

$154^o$

$=180^o - 154^o = 26^o$

1339231_1645991_ans_cd7618386ddd46a4a6a88657a81dbfd0.png

Find the values of the unknown x and y in the following diagram.

$\textbf{Step 1: Find the required angles applying suitable properties.}$

$\text{From the diagram, angle measuring x is an exterior angle of}$

$\triangle ABC$

$\text{We know, the exterior angle is the sum of two opposite interior angles. }$

$\therefore x=\angle ACB+\angle BAC$

$\Rightarrow x=60^\circ+50^\circ$

$\Rightarrow x=110^\circ.......(1)$

$\text{Again, in the diagram, angles measuring x and y are linear pair.}$

$\text{We know, linear pair are supplementary.}$

$\therefore x+y=180^\circ$

$\Rightarrow 110^\circ+y=180^\circ$

$[\textbf{From (1)}]$

$\Rightarrow y=70^\circ$

$\boldsymbol{\textbf{Hence, the values of x and y are } 110^\circ\textbf{ and } 70^\circ.}$

Find the supplement of the angle shown in the above figure.

Given angle is

$105^o$
Let supplementary angle of

$105^o$ be

$x$ .

We know that supplementary angles are those angles whose sum is

$180^o$ .

$\therefore \ 105^o+x=180^o$

$\Rightarrow x=180^o-105^o=75^o$

Hence, the supplementary angle of

$105^o$ is

$75^o$ .

1339378_1645989_ans_ec095a38a353437fa994a1aa718669fa.png

Find the complement of the angle represented in given figure.

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, angle is $20^o$ .

Then, the complement of $20^{0}=90^{0}-20^{0}=70^{0}$ .

1339402_1645982_ans_df1c81bd8f2545a3809246d67c3aaa02.png

Find the values of the unknown $x$ and $y$ in the following diagram.

$\textbf{Step 1: Find value of internal angles of triangle}$

$\text{Let the triangle be}$

$\triangle ABC$ .

$\text{Since,Vertically opposite angles are equal,then-}$

$\angle A = x$

$\angle B = x$

$\angle C = x = y$

$\text{In}$

$\triangle ABC,$

$\text{ By angle sum property-Sum of internal angles of a}$

$\Delta ABC$ is

$180^o$ .

$\Rightarrow \angle A + \angle B + \angle C = 180^o$

$\Rightarrow x + x+ x = 180^o$

$\Rightarrow 3x = 180^o$

$\therefore x = 60^o$

$\therefore y = x=60^o$

$\textbf{Therefore,value of x and y is $\mathbf{60^o}$}$

2181487_1645750_ans_6721b6e1bfc848d08c98307cd8332efd.png

Find the supplement of each of the following angles:

Supplementary angle

$= 180^o$ - given angle

$\Rightarrow$ Supplement of

$87^o = 180^o - 87^o =93^o$
1339377_1645990_ans_b22748f6bc654bc8b4db8a5561cb12b9.png

Find the values of the unknown $x$ and $y$ in the following diagram.

$\textbf{Step 1: Find the required angles applying suitable properties.}$

$\text{In the diagram, angles measuring y and}$

$90^\circ$

$\text{are vertically opposite angles.}$

$\text{We know, vertically opposite angles are equal .}$

$\therefore y=90^\circ......(1)$

$\text{We know, sum of interior angles of a triangles is }180^\circ.$

$\text{So, from the given triangle, }$

$\therefore x+x+y=180^\circ$

$\Rightarrow 2x+90^\circ=180^\circ$

$[\textbf{From (1)}]$

$\Rightarrow 2x=90^\circ$

$\Rightarrow x=45^\circ$

$\mathbf{\textbf{Hence, the values of x and y are } 45^\circ\ and \ 90^\circ\ respectively.}$

Two supplementary angles are in the ratio 4:Find the angles.

The given ratio of supplementary angles is

$4:5$

Let the angles be

$4x$ and

$5x$

$\Rightarrow$

$4x+5x=180^o$

$\Rightarrow$

$9x=180^o$

$\Rightarrow$

$x=\dfrac{180^o}{9}$

$\Rightarrow$

$x=20^o$

$\Rightarrow$

$4x=4\times 20^o=80^o$

$\Rightarrow$

$5x=5\times 20^o=100^o$

$\therefore$ Two supplementary angles are

$80^o$ and

$100^o.$

Write the supplement of the following angle: $54^{\circ}$

As we know for becoming supplementary angles, the sum of both the angles should be

$180^{\circ}$

Let the supplementary angle be

$x$

So,

$54^{\circ}+x=180^{\circ}$

$\Rightarrow x=126^{\circ}$

Write the complement (in degree) of the following angles: $77^{\circ}$

We know, two angles are complementary when they add up to $90^o$

Given, measure of one complementary angle is $77^{\circ}$

$\Rightarrow$ Measure of other complementary angle $=90^o-77^o$ $=13^o$

$\therefore$ Measure of a complementary angle of $77^{o}=13^o$

Write the complement (in degree) of the following angles: $20^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $20^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-20^o$ $=70^o.$

$\therefore$ Measure of a complementary angle of $20^{o}=70^o$ .

Write the supplement of the angle $132^{\circ}$

The measure of given angle is

$132^\circ$

Let the required angle be

$x$

Sum of supplementary angle is

$180^\circ$

$\therefore$

$x+132^\circ=180^\circ$

$\Rightarrow$

$x=180^\circ-132^\circ$

$\Rightarrow$

$x=48^\circ$

An angle is equal to $8$ times its complement. Determine its measure.

Let the required angle be

$x$ .

We know that the sum of complementary angles is

$90^{\circ}.$

Then, its complement

$=(90^{\circ}-x)$ .

Given, the angle is

$8$ times its complement.

Then,

$x=8(90^{\circ}-x)$

$\Rightarrow x=720^{\circ}-8x$

$\Rightarrow x+8x=720^{\circ}\\ \Rightarrow 9x=720^{\circ}\\ \Rightarrow x=\dfrac{720}{9}^{ \circ }=80^{\circ}$ .

Hence, the angle which is equal to

$8$ times its complement is

$80^{\circ}$ .

Write the complement (in degree) of the following angles: $30^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $30^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-30^o$ $=60^o.$

$\therefore$ Measure of a complementary angle of $30^{o}=60^o$ .

Can two angles be supplementary if both of them are: (i) acute (ii) obtuse (iii) right?

(i) No, because sum of two acute angles is less than

$180^o$ .
(ii) No, because sum of two obtuse angles is more than

$180^o$ ,
(iii) Yes, because sum of two right angles is

$180^o$ .

Write the complement (in degree) of the following angles: $35^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $35^o.$

$\Rightarrow$ Measure of other complementary angle $=90^o-35^o$ $=55^o.$

$\therefore$ Measure of a complementary angle of $35^{o}=55^o$ .

Write the complement (in degrees) of the following angles: $90^{\circ}$ .

We know, two angles are complementary when they add up to $90^o.$

Given, measure of one complementary angle is $90^{\circ}$ .

$\Rightarrow$ Measure of other complementary angle $=90^o-90^o$ $=0^o.$

$\therefore$ Measure of a complementary angle of $90^{o}=0^o$ .

In Fig., $l,m$ and $n$ are parallel lines intersected by traversal $p$ at $X,Y$ and $Z$ respectively. Find $\angle 1,\angle 2$ and $\angle 3$ .

1656324_1669328_ans_f8758209f59f4ba7bb4c13dee678d774.jpg

In FIG., show that $AB \parallel EF$ .

1656349_1669405_ans_91dd94d317e74d0d9b6a238593cbdfc6.jpg

In a $\Delta ABC,$ if $\angle A = 55^{\circ}, \angle B = 40^{\circ},$ find $\angle C.$ (value only)

In a

$\triangle ABC$ , by angle sum property, sum of angles of a triangle is

$180^{\circ}$

$\therefore \angle ABC + \angle BCA + \angle CAB = 180^{\circ}$

Given :

$\angle A=\angle CAB = 55^{\circ}$

and

$\angle B=\angle ABC = 40^{\circ}$ .

$\therefore 55^{\circ} + 40^{\circ} + \angle BCA = 180^{\circ}$

$\therefore \angle BCA = 180^{\circ} - 95^{\circ}$

$= 85^{\circ}$ .

Therefore,

$\angle BCA = \angle C=85^o$ .

1523797_1669545_ans_7424bfc95276411ca1dba2b1e39332c0.png

How many pairs of adjacent angles are formed when two lines intersect in a point?

$\textbf{Step -1: Drawing two lines intersecting at a point.}$

$\text{The following figure shows two lines intersecting at point}$

$O$

$\textbf{Step -2: Finding pairs of adjacent angles formed by the intersection of these lines}$

$\text{In the above figure, there are four pairs of adjacent angles as follows:}$

$\angle{AOC},\angle{COD}$

$\angle{COD},\angle{DOB}$

$\angle{DOB},\angle{BOA}$

$\angle{BOA},\angle{AOC}$

$\textbf{Final Answer: Hence, when two lines intersect in a point, four pairs of adjacent angles are formed.}$

In Fig., $AOC$ is a line, find $x$ .

$AOC$ is a line.

$\Rightarrow$

$\angle AOB+\angle BOC=180^o$ [ Linear pair ]

$\Rightarrow$

$70^o+2x=180^o$

$\Rightarrow$

$2x=180^o-70^o$

$\Rightarrow$

$2x=110^o$

$\Rightarrow$

$x=\dfrac{110^o}{2}$

$\Rightarrow$

$x=55^o$

Fill in the blanks in each of the following to make the statement true:
If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are _______

1656337_1669382_ans_d4ce3607b93e4560aba040a35bb3f53e.jpg

How many pairs of adjacent angles, in all, can you name in Fig.

Number of pair of adjacent angles, from the figure, are :

$\angle EOC$ and

$\angle DOC$

$\angle EOD$ and

$\angle DOB$

$\angle DOC$ and

$\angle COB$

$\angle EOD$ and

$\angle DOA$

$\angle DOC$ and

$\angle COA$

$\angle BOC$ and

$\angle BOA$

$\angle BOA$ and

$\angle BOD$

$\angle BOA$ and

$\angle BOE$

$\angle EOC$ and

$\angle COA$

$\angle EOC$ and

$\angle COB$

$\therefore$ There are

$10$ pairs of adjacent angles.

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.

Given: Two arms of an angle are perpendicular to the two arms of another angle.

To find : Angles are either equal or supplementary

Consider two angles

$\angle AOB$ and

$\angle ACB$

$OA\perp CA$ and

$OB \perp CB$

$\angle A + \angle O +\angle B+ \angle C=360^{\circ}$

$[$ angle sum property of a quadrilateral

$]$

$\implies 180^{\circ}+\angle O+\angle C=360^{\circ}$

$\implies \angle O +\angle C=360^{\circ}-180^{\circ}$

$\implies \angle AOB+\angle ACB=180^{\circ}$

$\therefore$ angles are supplementary

Now,

$\angle B+\angle ACB=180^{\circ}$ (sum of angles on the same side is supplementary)

$\implies 90^{\circ}+\angle ACB=180^{\circ}$

$\implies \angle ACB=90^{\circ}$ -------(1)

Now,

$\angle A+\angle A0B=180^{\circ}$ (sum of angles on the same side is supplementary)

$\implies 90^{\circ}+\angle A0B=180^{\circ}$

$\implies \angle A0B=90^{\circ}$ -------(2)

From (1) and (2)

$\angle AOB=\angle ACB=90^{\circ}$

2005723_1669410_ans_eb144c9367ed4941b4391298607c13e0.png

In Fig., ACB is a line such that $\angle DCA = 5x$ and $\angle DCB = 4x.$ Find the values of $\angle DCA$ and $\angle DCB.$

Here,

$ACB$ is a line.

$\Rightarrow$

$\angle DCA+\angle DCB=180^o$ [ Linear pair ]

$\Rightarrow$

$5x+4x=180^o$

$\Rightarrow$

$9x=180^o$

$\Rightarrow$

$x=\dfrac{180^o}{9}$

$\Rightarrow$

$x=20^o$

$\Rightarrow$

$\angle DCA=5x=5\times 20^o=100^o$

$\Rightarrow$

$\angle DCB=4x=4\times 20^o=80^o$

Classify the following angle as acute, obtuse, straight, right, zero and complete angle.
$118^o$ .

An angle whose measure is more than 90° but less than 180° is called an obtuse angle.

118° is greater than 90° and less than 180°. So it is obtuse angle.

Classify the following angle as acute, obtuse, straight, right, zero and complete angle.
$145^o$ .

An angle whose measure is more than 90° but less than 180° is called an obtuse angle.

145° is greater than 90° and less than 180°. So it is obtuse angle.

Fill in the blanks to make the following statements true.
Sum of the angles of a triangle is ______.

Sum of the angles of a triangle is

$180^{o}$

Compute the value of $x$ in the following figure:

$\triangle ABC,$

$\angle ACD+\angle ACB=180^o$ [Linear pair]

$\Rightarrow$

$110^\circ+\angle ACB=180^\circ$

$\therefore$

$\angle ACB=70^\circ$

Again,

$\Rightarrow$

$\angle ABE+\angle ABC=180^\circ$ [Linear pair]

$\Rightarrow$

$120^\circ+\angle ABC=180^\circ$

$\therefore$

$\angle ABC=60^\circ$

Now, in

$\triangle ABC,$

$\Rightarrow$

$\angle ABC+\angle ACB+\angle BAC=180^\circ$ [by angle sum property]

$\Rightarrow$

$60^\circ+70^\circ+x=180^\circ$

$\Rightarrow$

$130^\circ+x=180^\circ$

$\Rightarrow$

$x=50^\circ$

In a $\Delta PQR,$ If PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that $LN = MN.$

Using mid-point theorem,

we have

$MN \parallel PQ$ and

$MN = \dfrac{1}{2}PQ\Rightarrow MN = PL$

Similarly, we have

$LM = PN$

In triangles

$NML$ and

$LPN,$ we have

$MN = PL$

$LM = PN$

and,

$LN = NL$

So, by

$SSS$ congruence criterion, we obtain

$\Delta NML \cong \Delta LPN$

$\Rightarrow \angle MNL = \angle PLN$ and

$\angle MLN = \angle LNP$

$\Rightarrow \angle MNL = \angle LNP = \angle PLM = \angle MLN$

$\Rightarrow \angle PNM = \angle PLM$

$\therefore LN=MN$

1396950_1669861_ans_e77adf770abf4f8dafcb482091ca9dac.png

From figure, write all pairs of parallel lines.

The pairs of parallel lines are

$l || m, m || n$ and

$n || l$

Construct the following angles using set-squares: $75^{\circ}$

Place

$45^\circ$ set-square and make an angle of

$45^\circ$ by drawing the rays

$BD$ and

$BC$ as shown in the above figure.

Then, place the vertex of

$30^\circ$ of the set-square on the ray

$BD$ as shown in the figure and draw the ray

$BA$ .

The angle so formed is

$75^\circ$ .

Therefore,

$∠ABC = 75^\circ$

In Fig., line $n$ is a transversal to lines $l$ and $m$ . Identify the following:
Alternate and corresponding angles in Fig.

We know that the corresponding angles are the angles which occupy the same relative position at each intersection where a straight line crosses two others. Therefore, in the given figure, corresponding angles are as follows:

$∠EGB$ and

$∠GHD$

$∠HGB$ and

$∠FHD$

$∠EGA$ and

$∠GHC$

$∠AGH$ and

$∠CHF$

Also, we know that the alternate angles are two angles, formed when a line crosses two other lines, that lie on opposite sides of the transversal line and on opposite relative sides of the other lines. Therefore, in the given figure, alternate angles are as follows:

$∠EGB$ and

$∠CHF$

$∠HGB$ and

$∠CHG$

$∠EGA$ and

$∠FHD$

$∠AGH$ and

$∠GHD$

Classify the following angle as acute, obtuse, straight, right, zero and complete angle.
$165^o$ .

An angle whose measure is more than 90° but less than 180° is called an obtuse angle.

165° is greater than 90° and less than 180°. So it is obtuse angle.

Can a triangle have: All angles less than $60^{\circ}?$
Justify your answer in each case.

By angle sum property, we know, the sum of all angles of a triangle

$=180^0$ .

$\implies$

$60^0+60^0+60^0=180^0$ .

If all the these angles is less than

$60^0$ , then there sum together would not make

$180^0$ .

Say for example,

$30^0,40^0$ and

$50^0$ are

$3$ angles of the triangle.

$\therefore 30+40+50=120^0 < 180^0$ .

Therefore, it is not possible.

Hence, the answer is

$No$ .

In figure, write down each pair of vertically opposite angles.

As per the figure, there are 4 pair of vertically opposite angles.

$\angle1$ and $\angle 4$
$\angle2$ and $\angle 3$
$\angle5$ and $\angle 8$
$\angle 6$ and $\angle 7$

Write down each pair of adjacent angles shown in figure.

As per the definition, two angles are Adjacent when they have a common side and a common vertex.

According to the figure there are 4 pairs of adjacent angles.

$\angle$ DOC, $\angle$ COB
$\angle$ DOC, $\angle$ COA
$\angle$ COB, $\angle$ BOA and
$\angle$ AOB, $\angle$ BOD

In figure, name all the pairs of adjacent angles.

Adjacent angles

$\angle BAC \ and \angle CAD$

$\angle ABC \ and \angle ABE$

$\angle BCA \ and \angle BCF$

In figure, name all the pairs of adjacent angles.

As per the definition two angles are Adjacent when they have a common side and a common vertex.

As per the figure there are

$3$ pairs of adjacent angles

$\angle$

$ABE$ ,

$\angle$

$ABC$

$\angle$

$BAC$ ,

$\angle$

$CAD$ and

$\angle$

$ACB$ ,

$\angle$

$BCF$

Are the angles $1$ and $2$ given in adjacent angles?

no they are

$\text{alternate interior angles}$ not adjacent angles.

Classify the following angle as acute, obtuse, straight, right, zero, and complete angle.
$89.5^o$ .

An angle whose measure is more than

$0°$ but less than

$90°$ is called an acute angle.

Here,

$89.5°$ is less than

$90°$ . So it is an acute angle.

Classify the following angle as acute, obtuse, straight, right, zero and complete angle.
$180^o$ .

180 degrees is called as straight angle.

Classify the following angle as acute, obtuse, straight, right, zero and complete angle.
$75^o$ .

An angle whose measure is more than 0° but less than 90° is called an acute angle.

75° is less than 90°. So it is acute angle.

In figure, write down each linear pair.

As per the figure, there are

$8$ pair of linear angles.

$\angle1$ and $\angle 2$
$\angle1$ and $\angle 3$
$\angle4$ and $\angle 2$
$\angle 4$ and $\angle 3$
$\angle5$ and $\angle 6$
$\angle5$ and $\angle 7$
$\angle8$ and $\angle 6$ and
$\angle 8$ and $\angle 7$

One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?

let the measure of two linear pair angles be

$x$ and

$y$

We know that, the sum of angles in a linear pair is

$180^o$ .

$x + y= 180^o$

Given that, one of the angles is an obtuse angle.

obtuse angle

$+ y= 180^o$

$y= 180^o-$ obtuse angle

$y=$ acute angle

$\therefore$ other angle in a linear pair where one angle obtuse is an acute angle.

One of the angles forming a linear pair is an acute angle. What kind of angle is the other?

let the measure of two linear pair angles be

$x$ and

$y$

We know that the sum of angles in a linear pair is

$180^o$ .

$x + y= 180^o$

Given that, one of the angles is an acute angle.

obtuse angle

$+ y= 180^o$

$y= 180^o-$ acute angle

$y=$ obtuse angle

$\therefore$ another angle in a linear pair where one angle acute is an obtuse angle.

In figure, $A,\ B,\ C$ are collinear points and $\angle DBA=\angle EBA$ . Name two pairs of supplementary angles.

Supplementary angles are two angles whose sum is

$180^{\circ}$

According to question,

$\angle DBA=\angle EBA$

And

$line\ AC$ is a straight line.

So,

$\angle DBA+\angle DBC=180^\circ$

So, they satisfy condition for supplementary angles.

Similarly,

$\angle EBA+\angle EBC=180^\circ$

So, they also satisfy condition for supplementary angles.

$\therefore \angle DBA$ and

$\angle DBC$ as well as

$\angle EBA$ and

$\angle EBC$ are supplementary angles.

One of the angles forming a linear pair is a right angle. What can you say about the other angle?

We know that, two angles are said to be a linear pair of angles, if they are adjacent angles and their sum is

$180^o$ .

let the two angles be

$x$ and

$y$ , which formed linear pair.

Given, one of the angles is right angle.

Let,

$x = 90^o$

Now,

$x+ y = 180^o$

$\Rightarrow 90^o + y = 180^o$

$\Rightarrow y = 90^o$

Hence, the other angle is also a right angle.

If the complement of an angle is $28^o$ , then find the supplement of the angle.

Let the angle be

$x^o$

Given that

$x^o$ is the complement of an angle of

$28^o$ .

$x^o+28^o=90^o$

$x = 62 ^o$ .

Supplementary angle of

$62^o= 180^o-62^o$

$=118^o$

Find the complement of the following angle: $72^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $72^o$ .

Then, the complement of $72^{0}=90^{0}-72^{0}=18^{0}$ .

If two supplementary angles have equal measure, what is the measure of each angle?

Let the

$x$ be the measure of the two angles which are supplementary.

$x+x =180^o$

$2x=180^o$

$x=90^o$

$\therefore$ measure of an each angle =

$90^o$

In figure, name each linear pair and each pair of vertically opposite angles.

According to the figure,

Linear pair angles

$\angle 1$ and $\angle 4$
$\angle 1$ and $\angle 2$
$\angle 3$ and $\angle 4$
$\angle 3$ and $\angle 2$
$\angle 5$ and $\angle 8$
$\angle 5$ and $\angle 6$
$\angle 7$ and $\angle 8$
$\angle 7$ and $\angle 6$
$\angle 9$ and $\angle 12$
$\angle 9$ and $\angle 10$
$\angle 11$ and $\angle 10$
$\angle 11$ and $\angle 12$

Vertically Opposite angles

$\angle 1$ and $\angle 3$
$\angle 2$ and $\angle 4$
$\angle 5$ and $\angle 7$
$\angle 6$ and $\angle 8$
$\angle 9$ and $\angle 11$
$\angle 10$ and $\angle 12$

Name the four pairs of supplementary angles shown in figure.

According to the figure,

$\angle AOC$ and

$\angle AOD$

$\angle AOC$ and

$\angle COB$

$\angle AOD$ and

$\angle DOB$

$\angle DOB$ and

$\angle BOC$ are four pair of supplementary angles.

In figure, A, B, C are collinear points . Name two linear pairs.

According to the problem

$\angle ABD$ and

$\angle DBC$ ,

$\angle ABE$ and

$\angle EBC$ are linear pairs.

Can two acute angles form a linear pair?

No, two acute angles cannot form a linear pair. Because sum of two acute angles cannot be equal to

$180^o$ .

In figure, it being given that $\angle 1=65^o$ , find all other angles.

Given,

$\angle 1 = 65^o$

From the figure,

$\angle 1 = \angle 3$ &

$\angle 2 = \angle 4$ [∵ Vertically opposite angles are equal ]

$\angle 3 = 65^o$

and

$\angle 3$ and

$\angle 4$ ,

$\angle 1$ and

$\angle 4$ [∵ linear pair of angles]

$\angle 3 + \angle 4^o = 180^o$

$65^o + \angle 4^o = 180^o$

$\angle 4^o=115^o$

$\angle 2^o=115^o$ [∵ Vertically opposite angles ]

$\therefore \angle 1 = 65^o,\angle 2^o=115^o,\angle 3 = 65^o$ and

$\angle 4^o=115^o$

In figure, write all pairs of adjacent angles and all the linear pairs.

According to the figure,

There are

$4$ pairs of adjacent angles:

$\angle AOD$ &

$\angle DOC$

$\angle AOD$ &

$\angle DOB$

$\angle DOC$ &

$\angle COB$

$\angle COB$ &

$\angle COA$

There are

$2$ pairs of linear pair of angles

$\angle$ AOD and

$\angle DOB$

$\angle$ AOC and

$\angle COB$

Find the value of x in the following figure.

We know that, sum of all the angles in a straight line is

$180^o$

According to the given figure,

$3x^o+2x^o=180^o$

$5x^o=180^o$

$x^o=\dfrac{180}{5}^o$

$x=36^o$

Find the value of x in the following figure.

We know that, sum of all the angles in a straight line is

$180^o$

According to the given figure,

$60^o+x^o=180^o$

$x^o=180^o-60^o$

$x^o=120^o$

In figure, find $\angle x$ . Further find $\angle BOC, \angle COD$ and $\angle AOD$ .

We know that the sum of all the angles in a straight line is

$180^o$

$\angle BOC + \angle COD + \angle AOD = 180^o$ .

$x +20 +x + x+10 = 180^o$

$3x +30 = 180^o$

$3x = 150^o$

$x = 50^o$

$\angle COD = 50^o$

$\angle AOD = x+10 ^o$

$=60^o$

$\angle BOC = x+20 ^o$

$=70^o$

$\therefore$

$\angle COD = 50^o$ ,

$\angle AOD = 60^o$ and

$\angle BOC = 70^o$ .

Find the value of x in the following figure.

We know that, sum of all the angles in a straight line is

$180^o$

According to the given figure,

$35^o+x+ 60^o=180^o$

$95^o+x=180^o$

$x=180^o-95^o$

$\therefore\ x=85^o$

Hence, the value of

$x$ is

$85^o$ .

In figure, $OA$ and $OB$ are opposite rays, if $x=25^{\circ}$ , what is the value of $y^{\circ}$ ?

Given,

$x^{\circ}=25^{\circ}$

so,

$3x^{\circ}=3\times25^{\circ}=75^{\circ}$

We know that,

$\angle AOC, \angle BOC$ are linear pair of angles.

So,

$\angle AOC+\angle BOC=180^{\circ}$

$\Rightarrow (2y+5)^{\circ}+3x^{\circ}=180^{\circ}$

$\Rightarrow2y^{\circ}+5^{\circ}+75^{\circ}=180^{\circ}$

$\Rightarrow2y^{\circ}+80^{\circ}=180^{\circ}$

$\Rightarrow2y^{\circ}=180^{\circ}-80^{\circ}$

$\Rightarrow y^{\circ}= \dfrac{100^{\circ}}{2}$

$\Rightarrow y^{\circ}= 50^{\circ}$

Hence, the value of

$y^{\circ}$ is

$50^{\circ}$ .

In figure, OA and OB are opposite rays if $x=25^o$ , what is the value of y?

Given,

$x=25^o$ =>

$3x^o=75^o$

We know that

$\angle AOC, \angle BOC$ are linear pairs of angles.

$\therefore \angle AOC+\angle BOC=180^0$

$(2y+5)^o+3x^0=180$

$2y+5+75=180$ [ ∵

$3x^o=75^o$ ]

$2y+80=180$

$2y=180-80$

$y= \dfrac{100}{2}$

$\therefore y =50^\circ$

In figure, line n is a transversal to lines $l$ and $m$ . Identify the following alternate and corresponding angles in figure.

According to the figure,

List of alternate angles

$\angle AGF$ and

$\angle GHD$

$\angle CHG$ and

$\angle HGB$

$\angle AGE$ and

$\angle FHD$

$\angle EGB$ and

$\angle CHF$

List of corresponding angles

$\angle EGB$ and

$\angle GHD$

$\angle BGH$ and

$\angle DHF$

$\angle EGA$ and

$\angle GHC$

$\angle AGF$ and

$\angle CHF$

In figure, line $n$ is a transversal to lines $l$ and $m$ . Identify the following pairs of interior and exterior angles on the same side of the transversal in figure.

As per the figure,

$l||m$ and

$n$ is transversal,

Pair of interior angles on the same side of transversal :

$\angle a$ and

$\angle e$

$\angle d$ and

$\angle f$

Pair of interior angles on the same side of transversal :

$\angle b$ and

$\angle h$

$\angle c$ and

$\angle g$

In figure, $POS$ is a line, find $x$ .

Given,

$\angle POQ = 60^o$

$\angle QOR= 4x$

$\angle ROS = 40^o$

We know that, the sum of all the angles in a straight line is

$180^o$

According to the figure,

$\angle POQ+\angle QOR+ \angle ROS = 180^o$

$60^o+4x+40^o=180$

$100+4x=180^o$

$4x=80^o$

$x=20^o$

In figure, line $l || m$ and $n$ is a transversal. If $\angle 1=40^o$ , find all the angles and check that all corresponding angles and alternate angles are equal.

As per the diagram

$l||m$ and

$n$ is transversal,

$\angle 1 = 40^o$

We know sum of angles on a straight line is

$180^o$

$\angle 2 + 40 = 180^o$

$\angle 4 + 40 = 180^o$

$\angle 2 = 140^o$

$\angle 4 = 140^o$

We know that Vertically opposite angles are equal

$\angle 1 = \angle 3 = 40^o$

$\angle 5 = \angle 7$

$\angle 6 = \angle 8$

We know that alternate interior angle are equal

$\angle 3 = \angle 5 = 40^o$ [ From $$\angle 3 = 40^o$$]

$\angle 4 = \angle 6 = 140^o$ [ From

$\angle 4 = 140^o$ ]

Thus

$\angle 5 = \angle 7 = 40^o$ [ From

$\angle 5 = 40^o$ ]

$\angle 6 = \angle 8 = 140^o$ [ From

$\angle 6 = 140^o$ ]

$\therefore$ pair of corresponding angles are equal and alternate angles are equal

Corresponding angles

$\angle 1 =\angle 5 = 40^o$

$\angle 2 =\angle 6 = 140^o$

$\angle 3 =\angle 7 = 40^o$

$\angle 4 =\angle 8 = 140^o$

Alternate angles

$\angle 1 = \angle 7 = 40^o$

$\angle 2 = \angle 8 = 140^o$

$\angle 3 = \angle 5 = 40^o$

$\angle 4 = \angle 6 = 140^o$

In figure, line n is a transversal to lines $l$ and $m$ . Identify the following angles alternate to $\angle d$ and $\angle g$ and angles corresponding to angles $\angle f$ and $\angle h$ in figure.

Given

$l||m$ and

$n$ is transversal.

Alternate angles

$\angle d= \angle e$

$\angle g= \angle b$

Corresponding angles

$\angle f= \angle c$

$\angle h= \angle a$

In figure, $AOC$ is a line, find $x$ .

Given,

$\angle AOB = 70^o$

$\angle BOC = 2x^o$

We know that sum of all angles in a straight line is

$180^o$

$\angle AOB + \angle BOC = 180^o$

$70^o+2x= 180^o$

$2x= 110^o$

$x= 55^o$

How many pairs of adjacent angles, in all, can you name in figure?

Two angles are Adjacent when they have a common side and a common vertex (corner point) and don't overlap.

So,

$point\ O$ is a common vertex for all angles.

When

$OD$ is common side,

$\angle EOD\ and\ \angle AOD,\ \angle EOD\ and\ \angle BOD,\ \angle EOD\ and\ \angle COD$ are pair of adjacent angles.

Similarly, when

$OC$ is common side,

$\angle EOC\ and\ \angle AOC,\ \angle EOC\ and\ \angle BOC,\ \angle DOC\ and\ \angle AOC,\ \angle DOC\ and\ \angle BOC$ are pair of adjacent angles.

When

$OB$ is common side,

$\angle AOB\ and\ \angle EOB,\ \angle AOB\ and\ \angle BOD,\ \angle AOB\ and\ \angle COB$ are pair of adjacent angles.

In figure, $line\ XY\ ||\ line \ PQ$ and we have $\angle ALY=(2x-15)^o, \angle LMQ=(x+40)^o$ , find x.

We have

$\angle ALY=(2x-15)^0$ and

$\angle LMQ=(x+40)^0$

Also,

$line\ XY\ ||\ line \ PQ$

$\therefore$

$\angle XLM$ and

$\angle LMQ$ are corresponding angles.

and corresponding angles on parallel lines are congruent.

$\therefore \angle ALY=\angle LMQ$

$\therefore 2x-15=x+40\\$

$\Rightarrow 2x-x=40+15\\$

$\therefore x=55^0$

In figure, line $n$ is a transversal to lines $l$ and $m$ . Identify the following:
(a) angle alternate to $\angle PQR$ ,
(b)angle corresponding to $\angle RQF$ and
(c) angle alternate to $\angle PQE$ in figure.

$(a)$ Alternate angles are angles that are in opposite positions relative to a transversal intersecting two lines.

$line \ n$ is a transversal to lines

$l$ and

$m$

So,

$\angle ARQ$ is relatively opposite to angle

$PQR$ with respect to transversal

$n$ .

$\therefore \angle ARQ$ is angle alternate to

$\angle PQR$

$(b)$ Corresponding angles are the angles which occupy the same relative position at each intersection where a straight line crosses two others.

$line \ n$ is a transversal that is it crosses lines

$l$ and

$m$

So,

$\angle BRA$ is at relatively same position to angle

$RQF$ with respect to transversal

$n$ .

$\therefore \angle BRA$ is angle alternate to

$\angle RQF$

$(a)$ Alternate angles are angles that are in opposite positions relative to a transversal intersecting two lines.

$line \ n$ is a transversal to lines

$l$ and

$m$

So,

$\angle BRA$ is relatively opposite to angle

$PQE$ with respect to transversal

$n$ .

$\therefore \angle BRA$ is angle alternate to

$\angle PQE$

1615653_1677530_ans_e64b7ea690184f21bad8f63be7ab2b43.png

Name the triangle. Also, name the side opposite to $\angle B$ .

1616547_1677722_ans_bb69dd3b957640a288b706e7796399da.jpg

In the six concerned figure, AC, AD and AE are joined. Find $\angle FAB+\angle ABC+\angle BCD+\angle CDE+\angle DEF+\angle EFA$ .

1514338_1677815_ans_3d3688ff87204251b1aa48872086aad5.jpg

Find the value of $x$ for which the angles ${(2x-5)}^{o}$ and ${(x-10)}^{o}$ are the complementary angles.

It is given that

${(2x-5)}^{o}$ and

${(x-10)}^{o}$ are the complementary angles.

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then,

${(2x-5)}^{o}+{(x-10)}^{o}={90}^{o}$

$\implies$

$2x-{5}^{o}+x-{10}^{o}={90}^{o}$

$\implies$

$3x-{15}^{o}={90}^{o}$

$\implies$

$3x={15}^{o}+{90}^{o}$

$\implies$

$3x={105}^{o}$

$\implies$

$x=\dfrac{105^o}{3}$

$\implies$

$x={35}^{o}$ .

Therefore, the value of

$x$ for which the angles

${(2x-5)}^{o}$ and

${(x-10)}^{o}$ are the complementary angles is

${35}^{o}$ .

Find the supplement of the following angle:
$\dfrac{3}{5}$ of a right angle

We know that

$\dfrac{3}{5}$ of a right angle can be written as

$\dfrac{3}{5}\times 90^o={54}^{o}={180}^{o}-{54}^{o}={126}^{o}$

Find the measure of an angle that is equal to its supplement.

Consider the required angle as

${x}^{o}$

We know that the complement can be written as

${180}^{o}-{x}^{o}$

To find the measure of an angle is equal to its complement

We get

${x}^{o}={180}^{o}-{x}^{o}$

We can also write it as

$x+x=180$

so we get

$2x=180$

By division we get

$x=\dfrac{180}{2}$

${x}^{o}={90}^{o}$

Therefore the measure of an angle which is equal to its complement is

${90}^{o}$

Find the supplement of the following angle:
${42}^{o}$

We know that supplement of

${90}^{o}$ can be written as

${42}^{o}={180}^{o}-{42}^{o}={138}^{o}$

In the figure, what value of $x$ will make $AOB$ , a straight line?

We know that

$AOB$ will be straight line only if the adjacent angles form a linear pair

$\angle BOC+\angle AOC={ 180 }^{ o }$

${ x }+{ \left( 4x-36 \right) }^{ o }+{ \left( 3x+20 \right) }^{ o }={ 180 }^{ o }$

$\Rightarrow 4x-{ 36 }^{ o }+3x+{ 20 }^{ o }={ 180 }^{ o }$

$\Rightarrow 7x={ 180 }^{ o }-{ 20 }^{ o }+{ 36 }^{ o }$

$\Rightarrow 7x={ 196 }^{ o }\quad$

$\Rightarrow x={ 28 }^{ o }$

Therefore the value of

$x$ is

$28$

Define the following term:
Obtuse angle

Obtuse angle- An angle whose measure is more than

${90}^{o}$ but less than

${180}^{o}$ is called an obtuse angle.

Find the supplement of the following angle:
${124}^{o}$

We know that supplement of

${124}^{o}$ can be written as

${124}^{o}={180}^{o}-{124}^{o}={56}^{o}$

Define the following term:
Supplementary angles

Two angles are said to be supplementary if the sum of their measure is

${180}^{o}$

Define the following term:
Reflex angle

An angle whose measure is more than

${180}^{o}$ but less than

${360}^{o}$ is called reflex angle.

For example,

$280^o, 181^o,$ etc. are reflex angles as they are greater than

$180^o$ and less than

$360^o$ .

Two adjacent angles on a straight line are in the ratio $5:4$ . Find the measure of each one of these angles.

Consider the two adjacent angles as

$5x$ and

$4x$

We know that the two adjacent angles form a linear pair

So it can be written as

$5x+4x={ 180 }^{ o }\Rightarrow 9x={ 180 }^{ o }$

$x={20}^{o}$

substituting the value of

$x$ in two adjacent angles

$5x=5(20)={100}^{o}$

$4x=4(20)={80}^{o}$

Therefore the measure of each one these angles is

${100}^{o}$ and

${80}^{o}$

In $\triangle PQR$ , if $\angle P - \angle Q = 42^{\circ}$ and $\angle Q - \angle R = 21^{\circ}$ , find $\angle P, \angle Q$ and $\angle R$ .

It is given that

$\angle P - \angle Q = 42^{\circ}$

It can be written as

$\angle P = 42^{\circ} + \angle Q$

We know that the sum of all the angles in a triangle is

$180^{\circ}$ .

So we can write it as

$\angle P + \angle Q + \angle R = 180^{\circ}$

By substituting

$\angle P = 42^{\circ} + \angle Q$ in the above equation

$42^{\circ} + \angle Q + \angle Q + \angle R = 180^{\circ}$

On further calculation

$42^{\circ} + 2 \angle Q + \angle R = 180^{\circ}$

$2 \angle Q + \angle R = 180^{\circ} - 42^{\circ}$

By subtraction we get

$2 \angle Q + \angle R = 138^{\circ} . (i)$

It is given that

$\angle Q - \angle R = 21^{\circ}$

It can be written as

$\angle R = \angle Q - 21^{\circ}$

By substituting the value of

$\angle R$ in equation (i)

$2 \angle Q + \angle Q - 21^{\circ} = 138^{\circ}$

On further calculation

$3 \angle Q - 21^{\circ} = 138^{\circ}$

$3 \angle Q = 138^{\circ} + 21^{\circ}$

By addition

$3 \angle Q = 159^{\circ}$

By division

$\angle Q = 159/3$

$\angle Q = 53^{\circ}$

By substituting

$\angle Q = 53^{\circ}$ in

$\angle P = 42^{\circ} + \angle Q$

So we get

$\angle P = 42^{\circ} + 53^{\circ}$

By addition

$\angle P = 95^{\circ}$

By substituting

$\angle Q$ in

$\angle Q - \angle R = 21^{\circ}$

$53^{\circ} - \angle R = 21^{\circ}$

On further calculation

$\angle R = 53^{\circ} - 21^{\circ}$

By subtraction

$\angle R = 32^{\circ}$

Therefore,

$\angle P = 95^{\circ}, \angle Q = 53^{\circ}$ and

$\angle R = 32^{\circ}$ .

In the given figure, $AB$ and $CD$ are straight lines through the

centre $O$ of a circle. If $\angle AOC=80^o$ and $\angle CDE=40^o$ , find $\angle ABC$

We know that

$\angle AOC$ and

$\angle BOC$ from a linear pair

It can be written as

$\angle BOC =180^o -80^o$

By subtraction

$\angle BOC =100^o$

Using the angle sum property

$\angle ABC +\angle BOC +\angle DCE =180^o$

By substituting the values

$\angle ABC +100^o +50^o +=180^o$

On further calculation

$\angle ABC=180^o-100^o -50^o$

By subtraction

$\angle ABC =180^o -150^o$

So we get

$\angle ABC =30^o$

In the given figure, $AB$ and $CD$ are straight lines through the centre $O$ of a circle. If $\angle AOC=80^o$ and $\angle CDE=40^o$ , find $\angle DCE$

From the figure we know that

$\angle CED =90^o$

Consider

$\triangle CED$

Using the angle sun property

$\angle CED +\angle EDC +\angle DCE =180^o$

By substituting the values

$90^o +40^o +\angle DEC =180^o$

On further calculation

$\angle DCE=180^o-90^o -40^o$

By subtraction

$\angle DCE =180^o -130^o$

so we get

$\angle DCE =50^o$

In the given figure, $O$ is the centre of the circle. If $\angle PBC=25^o$ and $\angle APB=110^o$ , find the value of $\angle ADB$ .

We know that

$\angle ACB=\angle PCB$

$\triangle PCB$

Using the angle sum property

$\angle PCB+\angle BPC+\angle PBC=180^o$

We know that

$\angle APB$ and

$\angle BPC$ are linear pair

By substituting the values

$\angle PCB +(180^o -110^o)+25^o =180^o$

On further calculation

$\angle PBC +70^o+25^o =180^o$

$\angle PCB +95^o =180^o$

By subtraction

$\angle PCB =180^o -95^o$

So we get

$\angle PCB =85^o$

We know that the angles in the same segment of a circle equal

$\angle ADB=\angle ACB =85^o$

Therefore, the value of

$\angle ADB$ is

$85^o$

1566205_1715401_ans_23eea5a4f8da4e64aa8381719cfb953c.png

In the given figure, $AB\parallel CD$ . Find the values of $x,y$ and $z$ .

It is given that

$AB\parallel CD$ and

$EF$ is a transversal

from the fiture we know that

$\angle AEF$ and

$\angle EFG$ are alternate angles

so we get

$\angle AEF=\angle EFG={ 75 }^{ o }$

$\angle EFG=y={75}^{o}$

from the figure we know that

$\angle EFC$ and

$\angle EFG$ form a linear pair of angles

so we get

$\angle EFC+\angle EFG={ 180 }^{ o }$

it can also be written as

$x+y={ 180 }^{ o }$

by substituting the value of

$y$ we get

$x+{75}^{o}={ 180 }^{ o }$

$x={105}^{o}$

from the figure based on the exterior angle property it can be written as

$\angle EGD=\angle EFG+\angle FEG$

by substituting the values in the above equation we get

${125}^{o}=y+z$

${125}^{o}={75}^{o}+z$

$z={50}^{o}$

therefore the values of

$x,y,z$ are

${105}^{o},{75}^{o},{50}^{o}$

In the given figure, $O$ is the centre of the circle. If $\angle ABD=35^o$ and $\angle BAC=70^o$ , find $\angle ACB$ .

We know that

$BD$ is the diameter of the circle

Angle in a semicircle is a right angle

$\angle BAD=90^o$

Consider

$\triangle BAD$

Using the angle sum property

$\angle ADB +\angle BAD +\angle ABD=180^o$

By substituting the values

$\angle ADB +90^o +35^o =180^o$

On further calculation

$\angle ADB =180^o - 90^o -35^o$

By subtraction

$\angle ADB =180^o -125^o$

So we get

$\angle ADB =55^o$

We know that the angle in the same segment of a circle are equal

$\angle ACB =\angle ADB=55^o$

So we get

$\angle ACB =55^o$

Therefore,

$\angle ACB =55^o$

In the figure, $ABCD$ is a square and $\triangle EDC$ is an equilateral triangle. Prove that
(i) $AE=BE$
(ii) $\angle DAE={15}^{o}$

(i) From the figure we know that

$\triangle EDC$ is an equilateral triangle

so we get

$\angle EDC=\angle ECD={60}^{o}$

We know that

$ABCD$ is a square

so we get

$\angle CDA=\angle DCB={90}^{o}$

consider

$\triangle EDA$

we get

$\angle EDA={60}^{o}+{90}^{o}$

so we get

$\angle EDA={150}^{o}....(1)$

Consider

$\triangle ECB$

we get

$\angle ECB=\angle ECD+\angle DCB$

by substituting the values in the above equation

$\angle ECB={60}^{o}+{90}^{o}$

we get

$\angle ECB={150}^{o}$

so we get

$\angle EDA=\angle ECB....(ii)$

consider

$\triangle EDA$ and

$\triangle ECB$

$ED=EC$ (sides of an equilateral triangle)

$\angle EDA=\angle ECB$ (From

$(ii)$ )

$DA=CB$ (sides of square)

by SAS congruence criterion

$\triangle EDA\cong \triangle ECB$

$AE=BE$

(ii) consider

$\triangle EDA$

we know that

$ED=DA$

from the figure we know that the base angles are equal

$\angle DEA=\angle DAE$

based on equation (i) we get

$\angle EDA={150}^{o}$

by angle sum property

$\angle EDA+\angle DAE+\angle DEA={180}^{o}$

by substituting the values we get

${150}^{o}+\angle DAE+\angle DEA={ 180 }^{ o }$

we know that

$\angle DEA=\angle DAE$

so we get

${150}^{o}+\angle DAE+\angle DAE={ 180 }^{ o }$

$2\angle DEA={ 180 }^{ o }-{150}^{o}$

$\angle DAE={15}^{o}$

In $\triangle ABC$ , if $\angle A + \angle B = 108^{\circ}$ and $\angle B + \angle C = 130^{\circ}$ , find $\angle A, \angle B$ and $\angle C$ .

It is given that

$\angle A + \angle B = 108^{\circ} \dots (1)$

We know that the sum of all the angles in a triangle is

$180^{\circ}$ .

So we can write it as

$\angle A + \angle B + \angle C = 180^{\circ}$

By substituting

$\angle A + \angle B = 108^{\circ}$ in the above equation

$108^{\circ} + \angle C = 180^{\circ}$

On further calculation

$\angle C = 180^{\circ} - 108^{\circ}$

By subtraction

$\angle C = 72^{\circ}$

It is given that

$\angle B + \angle C = 130^{\circ}$

By substituting the value of

$\angle C$

$\angle B + 72^{\circ} = 130^{\circ}$

On further calculation

$\angle B = 130^{\circ} - 72^{\circ}$

By subtraction

$B = 58^{\circ}$

By substituting

$\angle B = 58^{\circ}$ in equation (1)

So we get

$\angle A + \angle B = 108^{\circ}$

$\angle A + 58^{\circ} = 108^{\circ}$

On further calculation

$\angle A = 108^{\circ} - 58^{\circ}$

By subtraction

$\angle A = 50^{\circ}$

Therefore,

$\angle A = 50^{\circ}, \angle B = 58^{\circ}$ and

$\angle C = 72^{\circ}$ .

In $\triangle ABC$ , if $\angle A + \angle B = 125^{\circ}$ and $\angle A + \angle C = 113^{\circ}$ , find $\angle A, \angle B$ and $\angle C$ .

It is given that

$\angle A + \angle B = 125^{\circ} \dots (1)$

We know that the sum of all the angles in a triangle is

$180^{\circ}$ .

So we can write it as

$\angle A + \angle B + \angle C = 180^{\circ}$

By substituting

$\angle A + \angle B = 125^{\circ}$ in the above equation

$125^{\circ} + \angle C = 180^{\circ}$

On further calculation

$\angle C = 180^{\circ} - 125^{\circ}$

By subtraction

$\angle C = 55^{\circ}$

It is given that

$\angle A + \angle C = 113^{\circ}$

By substituting the value of

$\angle C$

$\angle A + 55^{\circ} = 113^{\circ}$

On further calculation

$\angle A = 113^{\circ} - 55^{\circ}$

By subtraction

$\angle A = 58^{\circ}$

By substituting

$\angle A = 58^{\circ}$ in equation (1)

So we get

$\angle A + \angle B = 125^{\circ}$

$58^{\circ} + \angle B= 125^{\circ}$

On further calculation

$\angle B = 125^{\circ} - 58^{\circ}$

By subtraction

$\angle B = 67^{\circ}$

Therefore,

$\angle A = 58^{\circ}, \angle B = 67^{\circ}$ and

$\angle C = 55^{\circ}$ .

In the adjoining figure, $DE$ is a chord parallel to a diameter $AC$ of the circle with centre $O$ . If $\angle CBD=60^o$ , calculate $\angle CDE$ .

We know that the angles in the same segment of a circle are equal

From the figure we know that

$\angle CAD$ and

$\angle CBD$ are in the segment

$CD$

$\angle CAD =\angle CBD =60^o$

An angle in a semi-circle is a right angle

So we get

$\angle ADC =90^o$

Using the angle sum property

$\angle ACD +\angle ADC+\angle CAD =180^o$

By substituting the values

$\angle ACD +90^o +60^o =180^o$

$\Rightarrow \angle ACD =180^o -90^o -60^o$

$\Rightarrow ACD =180^o -150^o$

$\Rightarrow \angle ACD =30^o$

We know that

$AC || DE$ and

$CD$ is a transversal

From the figure we know that

$\angle ACD$ and

$\angle CDE$ are alternate angles

So we get

$\angle CDE =\angle ACD =30^o$

Therefore,

$\angle CDE =30^o$

State the type of each of the following angles:

$\angle AOB$ is an obtuse angle since measure is more than

$90^{o}$ but less than

$180^{o}$

Calculate the value of $x$ in each of the following figures.

We know that the sum of all the angles in

$\triangle ABE$ is

$180^{\circ}$ .

$\angle A + \angle B + \angle E = 180^{\circ}$

By substituting the values of

$\angle A$ and

$\angle B$ in the above equation:

$75^{\circ} + 65^{\circ} + \angle E = 180^{\circ}$

$\angle E = 180^{\circ} - 75^{\circ} - 65^{\circ}$

By subtraction

$\angle E = 180^{\circ} - 140^{\circ}$

$\angle E = 40^{\circ}$

From the figure we know that,

$\angle CED$ and

$\angle AEB$ are vertically opposite angles

So, we get:

$\angle CED = \angle AEB = 40^{\circ}$

We know that the sum of all the angles in

$\triangle CED$ is

$180^{\circ}$ .

$\angle C + \angle E + \angle D = 180^{\circ}$

By substituting the values:

$110^{\circ} + 40^{\circ} + x = 180^{\circ}$

On further calculation:

$x = 180^{\circ} - 110^{\circ} + 40^{\circ}$

$x = 180^{\circ} - 150^{\circ}$

$x = 30^{\circ}$ .

In the given figure, $O$ is the centre of a circle in which $\angle OAB=20^o$ and $\angle OCB=55^o$ . Find
$\angle BOC$

We know that

$OB=OC$ which is the radius

The base angles of an isosceles triangle are equal

So we get

$\angle OBC =\angle OCB =55^o$

$\triangle BOC$

Using the angle sum property

$\angle BOC + \angle OCB +\angle OBC =180^o$

By substituting values

$\angle BOC +55^o +55^o =180^o$

On further calculation

$\angle BOC =180^o -110^o$

So we get

$\angle BOC =70^o$

Two angles of a triangle are equal and the third angle is greater than each one of them by $18^{\circ}$ . Find the angles.

Consider

$\angle A$ and

$\angle B$ in a triangle is

$x^{\circ}$

We know that the sum of all the angles in a triangle is

$180^{\circ}$ .

So we can write it as

$\angle A + \angle B + \angle C = 180^{\circ}$

By substituting the values

$x^{\circ} + x^{\circ} + \angle C = 180^{\circ}$

By addition

$2x^{\circ} + \angle C = 180^{\circ} .. (1)$

According to the question we get

$\angle C = x^{\circ} + 18^{\circ} . (2)$

By substituting (2) in (1) we get

$2x^{\circ} + x^{\circ} + 18^{\circ} = 180^{\circ}$

On further calculation

$3x^{\circ} + 18^{\circ} = 180^{\circ}$

By subtraction

$3x^{\circ} = 180^{\circ} - 18^{\circ}$

$3x^{\circ} = 162^{\circ}$

By division

$x^{\circ} = 162/3$

$x^{\circ} = 54^{\circ}$

By substituting the values of

$x$

$\angle A = \angle B = 54^{\circ}$

$\angle C = 54^{\circ} + 18^{\circ} = 72^{\circ}$

Therefore,

$\angle A = 54^{\circ}, \angle B = 54^{\circ}$ and

$\angle C = 72^{\circ}$ .

State the type of each of the given angle:

Here,

$\angle POQ$ is reflex angle, since its measure is more than

$180^{o}$ but less than

$360^{o}$ .

Hence, the given angle is a reflex angle.

In a $\triangle ABC$ , it is given that $\angle A : \angle B : \angle C = 3 : 2 : 1$ and $CD \perp AC$ . Find $\angle ECD$ .

In a

$\triangle ABC$ , it is given that

$\angle A : \angle B : \angle C = 3 : 2 : 1$

It can also be written as

$\angle A = 3x, \angle B = 2x$ and

$\angle C = x$

We know that the sum of all the angles in triangle

$ABC$ is

$180^{\circ}$ .

$\angle A + \angle B + \angle C = 180^{\circ}$

By substituting the values we get

$3x + 2x + x = 180^{\circ}$

By addition

$6x = 180^{\circ}$

By division

$x = 180/6x = 30^{\circ}$

Now by substituting the value of

$x$ we get

$\angle A = 3x = 3 (30^{\circ}) = 90^{\circ}$

$\angle B = 2x = 2 (30^{\circ}) = 60^{\circ}$

$\angle C = x = 30^{\circ}$

We know that in the triangle

$ABC$ exterior angle is equal to the sum of two opposite interior angles

So we can write it as

$\angle ACE = \angle A + \angle B$

By substituting the values we get

$\angle ACE = 90^{\circ} + 60^{\circ}$

By addition

$\angle ACE = 150^{\circ}$

We know that

$\angle ACE$ can be written as

$\angle ACD + \angle ECD$

So we can write it as

$\angle ACE = \angle ACD + \angle ECD$

By substituting the values we get

$150^{\circ} = 90^{\circ} + \angle ECD$

It is given that

$CD \perp AC$ so

$\angle ACD = 90^{\circ}$

On further calculation

$\angle ECD = 150^{\circ} - 90^{\circ}$

By subtraction

$\angle ECD = 60^{\circ}$

Therefore,

$\angle ECD = 60^{\circ}$ .

Classify the angles whose magnitudes are given below:
$30^{o}$

Given angle is

$30^{o}$

We know that, an angle whose measure is less than

$90^o$ is called an acute angle.

Since,

$30^o<90^o$ , the given angle is an acute angle.

State the type of each of the following angles:

$\angle HOG$ is an acute angle since its measure is less than

$90^{o}$

In the given figure, $O$ is the centre of a circle in which $\angle OAB=20^o$ and $\angle OCB=55^o$ . Find
$\angle AOC$

We know that

$OA=OB$ which is the radius

The base angles of an isosceles triangle are equal

So we get

$\angle OBA =\angle OAB =20^o$

$\triangle AOB$

Using the angle sum property

$\angle AOB + \angle OAB +\angle OBA =180^o$

By substituting values

$\angle AOB +20^o +20^o =180^o$

On further calculation

$\angle AOB =180^o -20^o -20^o$

By subtraction

$\angle AOB =180^o -40^o$

So we get

$\angle AOB =140^o$

We know that

$\angle AOC =\angle AOB -\angle BOC$

By substituting the values

$\angle AOC =140^o -70^o$

So we get

$\angle AOC =70^o$

State the type of each of the following angles:

$\angle FOE$ is a straight angle whose measure is

$180^{o}$

Classify the angles whose magnitudes are given below:
$1\dfrac{1}{2}$ right angles?

$(1\times 1/2)$ right angle measure

$=(3/2\times 90^{o})=135^{o}$

Classify the angles whose magnitudes are given below:
$128^{o}$

$128^{o}$ is an obtuse angle since it is more than

$90^{o}$ but less than

$180^{o}$

Classify the angles whose magnitudes are given below:
$(90.5)^{o}$

$(90.5)^{o}$ is an obtuse angle since it is more than

$90^{o}$

Classify the angles whose magnitudes are given below:
$(38.3)^{o}$

$(38.3)^{o}$ is an acute angle since it is less than

$90^{o}$

Classify the angles whose magnitudes are given below:
$80^{o}$

$80^{o}$ is an acute angle since it is less than

$90^{o}$

Classify the angles whose magnitudes are given below:
$91^{o}$

$91^{o}$ is an obtuse angle as it is more than

$90^{o}$ and less than

$180^{o}$

Classify the angles whose magnitudes are given below:
$179^{o}$

$179^{o}$ is an obtuse angle since it is more than

$90^{o}$ and less than

$180^{o}$

Classify the angles whose magnitudes are given below:
$2/3$ right angle?

$(2/3\times 90^{o})$ measure

$=2\times 30^{o}=60^{o}$

Classify the angles whose magnitudes are given below:
$15^{o}$

$15^{o}$ is an acute angle since it is less than

$90^{o}$

Classify the angle whose magnitude is given below:
$181^{o}$

$A$ and

$B$ be two obtuse angles.

Then by definition,

$90^\circ<A<180^\circ$ and

$90^\circ<B<180^\circ$ .

Adding the two equations, we get,

$\therefore 90^\circ+90^\circ<A+B<180^\circ+180^\circ$

$\Rightarrow 180^\circ<A+B<360^\circ\qquad...(i)$

(a) From equation (i),

$180^\circ<A+B<360^\circ$

i.e.,

$A+B$ is an obtuse angle.

(b) From equation (i),

$A+B<360^\circ$

Hence,

$A+B$ can not be a complete angle.

In figure
(a) Name any four angles that appear to be acute angles
(b) Name any two angles that appear to be obtuse angles.

In the figure,

(a) Acute angle are

$\angle EAD,\angle EDA,\angle ABE,\angle EBC\quad$
(b) Obtuse angle are

$\angle BCD,\angle BAD$

Can we have two acute angles whose sum is
(a) An acute angle? Why or why not?
(b) A right angle? Why or why not?
(c) An obtuse angle? Why or why not?
(d) A straight angle? Why or why not?
(e) A reflex angle? Why or why not?

Let,

$A$ and

$B$ be two acute angles.

Then by definition,

$0^\circ<A<90^\circ$ and

$0^\circ<B<90^\circ$

Adding the two equations, we get

$\therefore 0^\circ+0^\circ<A+B<90^\circ+90^\circ$

$\Rightarrow 0^\circ<A+B<180^\circ\qquad...(i)$

Therefore,

(a) We can have two acute angles whose sum is an acute angle.

(b) We can have two acute angles whose sum is a right angle.

(d) We can have two acute angles whose sum is an obtuse angle.

(e) We can't not have two acute angles whose sum is a straight angle.

(f) We can't have two acute angles whose sum is an reflex angle.

Number of angles less than ${180}^{o}$ in figure is and their names are _.

$\textbf{Finding the angles less than 180°.}$

Angles less than 180° are as follows,

$∠OBA,∠OAB,∠AOB,∠OAC,∠OCA,∠AOC,∠ODC,∠OCD,∠DOC,∠BOD,∠BAC and ACD.$

$\textbf{Hence, there are 12 such angles which are less than 180°.}$

An angle greater than ${180}^{o}$ and less than a complete angle is called ______.

An angle greater than

$180^o$ and less than a complete angle is called

$\underline{reflex \ angle }$

For example::

The supplement of a right angles is always ________ angle.

The supplement of a right angles is always Right angle.

The supplement of an obtuse angle is always ______ angle.

The supplement of an obtuse angle is always acute angle.

If sum of measures of two angles is $90^{o}$ , then the angles are ______.

Complementary angles are two angles whose sum is

$90$ degrees.

Alternate interior angles are on the _______ side of the transversal.

Consider,

$m$ and

$n$ are the straight lines and transversal

$l$ intersects both the lines

$m$ and

$n$ .

Therefore, the angle

$a$ and

$b$ are alternate angles because of their position, that is, in the opposite side of the transversal and they are both lying on inside of a region covered by line

$m$ and

$n$ .

So, they are alternate interior angles.

Hence, we can say that alternate interior angles are on the opposite sides of the transversal.

1958524_1789947_ans_832dd3d17ba24b9baf3453961600eeb5.png

Alternate interior angles have one common ______.

Consider,

$m$ and

$n$ are the straight lines and transversal

$l$ intersects both the lines

$m$ and

$n$ .

Therefore, the angle

$a$ and

$b$ are alternate angles because of their position, that is, in the opposite side of the transversal and they are both lying on inside of a region covered by line

$m$ and

$n$ .

So, they are alternate interior angles.

Here, clearly the alternate interior angles are formed on the either side of a line segment

Hence, we can say that alternate interior angles have one common arm.

1958529_1789937_ans_b46e463ef9024160a031da46bdde56a2.png

State True or False:
In a pair of complementary angles, each angle cannot be more than $90^{o}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the each angles are more than $90^o$ ,

Let these angles be $90^o+x$ and $90^o+y$ .

Then, their sum $=90^o+x+90^o+y=180^o+x+y>90^o$ .

Therefore, they are not complementary.

Hence, in a pair of complementary angles, each angle cannot be more than

$90^{o}$ .

That is, the given statement is true.

Two complementary angles differ by $20^o$ . Find the angles.

Let $x$ be the one angle.

So, the complement of $x=90^o -x$ .

Given, two complementary angles differ by $20^o$ .

$\Rightarrow x-(90^o -x)=20^o$

$\Rightarrow x-90^o +x=20^o$

$\Rightarrow \ 2x=20^o+90^o$

$\Rightarrow \ x=\dfrac {110^o}{2}$

$\Rightarrow \ x=55^o$ .

Therefore, complementary of $x$ ,

$=90^o -x$

$=90^o -55^o$

$=35^o$ .

Hence, $35^o$ and $55^o$ are the complementary angles.

In each of the above figure write, if any, $(i)$ each pair of vertically opposite angles, and $(ii)$ each linear pair.

Here, no pair of vertically opposite angles are present.

$\angle ABD$ and

$\angle DBC; \angle ABE$ and

$\angle CBE$ are linear pairs.

Name the pairs of supplementary angles in the above figure:

1900260_1791017_ans_2afe364d814d436aa4f918fc6ee47649.jpg

Write down each pair of adjacent angles shown in the above figures:

1900208_1790522_ans_5455103eddb64893a15a23ce7e204e6e.jpg

Name the pairs of supplementary angles in the above figure:

Three pairs of supplementary angles are

$\angle 1$ and

$\angle 2, \angle 3$ and

$\angle 5$ and

$\angle 6$ .

In the above figure write, if any, $(i)$ each pair of vertically opposite angles, and $(ii)$ each linear pair.

$\angle POR$ and

$\angle QOS; \ \angle ROQ$ and

$\angle POS$ are two pairs of vertically opposite angles.

$\angle POR$ and

$\angle ROQ; \ \angle ROQ$ and

$\angle OOS; \ angle OOS$ and

$\angle SOP; \ \angle SOP$ and

$\angle POR; \ \angle ROT$ and

$\angle TOS; \ \angle OOT$ and

$POT$ are linear pairs of angles.

Two angles are making a linear pair. If one of them is one-third of the other, find the angles.

Let one angle be

$y$ .

As we know, the linear pair consists of two angles.

$\therefore 180^o-y$

$ACQ$ ,

$\Rightarrow y=\dfrac{1}{3}(180^o-y)$

$\Rightarrow 3y=180^o-y$

$\Rightarrow 3y=180^o$

$\Rightarrow y=\dfrac{180^o}{4}$

$\Rightarrow y=45^o$

Therefore, one angle is

$45$ degrees and other is

$135$ degrees.

$\Rightarrow 180^o-45^o=135^o$

In Fig. $P,Q$ and $R$ are collinear points and $TQ \bot PR,$
Name:
Two pair of supplementary angles.

Two pairs of supplementary angles are

$\angle PQT$ and

$TOR$ and

$\angle SQR$ and

$\angle PQS$ .

Can two acute angles form a pair of supplementary angles? Give reason support of your answer.

No, if both are acute angles then it cannot be a supplementary angles.
Example:

$89+89 \ne 180$

In the figure, $P,Q$ and $R$ are collinear points and $TQ \bot PR,$
name: A pair of complementary angles.

Complementary angles are those angles which adds to

$90^o$ .

Here,

$\angle TQS+\angle RQS=90^o$ .

Hence,

$\angle TQS$ and

$\angle RQS$ are a pair of complementary angles.

Two lines $AB$ and $CD$ intersect at $O$ Fig. Write all the pairs of adjacent angles by taking angles $1,\ 2,\ 3,\$ and $4$ only.

There are four pair of adjacent angles are formed:

$\angle 1$ and

$\angle 2, \angle 1$ and

$\angle 4, \angle 2$ and

$\angle 3, \angle 3$ and

$\angle 4$ .

Measure (in degree) of two supplementary angles are consecutive odd integers. Find the angles.

Let one angle be

$2y+1$ and other angle be

$2y+3$

As we know sum of supplementary angle is

$180$ degrees.

$ACQ,$

$\Rightarrow 2y+1+2y+3=180^o$

$\Rightarrow 4y+4=180^o$

$\Rightarrow 4y=180^o - 4^o$

$\Rightarrow y=\dfrac{176^o}{4}$

$\Rightarrow y=4^o$

$\therefore 2 \times y+1=3 \times 44^o+1=89^o$

$\Rightarrow$ and

$2y+3=2 times 44^o +3=91^o$

Therefore, one angle is

$89$ degrees and other is

$91$ degrees.

What is the type of other angle of a linear pair if one of its angle is obtuse?

In linear pair, if one angle is obtuse then the other will be an acute angle.

What is the type of other angle of a linear pair if one of its angle is acute?

In linear pair, if one angle is acute then the other will be an obtuse angle.

Measures (in degree) of two complementary angles are two consecutive even integers. Find the angles.

Let one angle be

$2y$ and other angle be

$2y+2^o$ .

As we know, the sum of complementary angles is

$90^o$

Then,

$\Rightarrow 2y+(2y+2^o)=90^o$

$\Rightarrow 2y+2y+2^o=90^o$

$\Rightarrow 4y=90^o-2^o$

$\Rightarrow 4y=88^o$

$\Rightarrow y=\dfrac{88^o}{4}$

$\Rightarrow y=22^o$ .

$\therefore 2y=2 \times 22^o =44^o$

and

$2y+2=44^o+2=46^o$ .

Therefore, one angle is

$44^o$ and other angle is

$46^o$ .

If the complementary of an angle is $62^o$ , then find its supplement.

Let

$y$ be the angle then its complement will be

$90^o-y$ .

$\Rightarrow 90^o-y=62^o$

$\Rightarrow y=90^o-62^o$

$\Rightarrow y=28^o$

Therefore, the supplement of

$y$

$\Rightarrow 180^o-28^o$

$=152^o$

Locate the points $P(3, 4),\ Q(1, 0),\ R (0, 4)$ and $S(4, 1)$ on a graph sheet and write the coordinates of the point of intersection of line segments $PQ$ and $RS$ .

The given points

$P, Q, R$ and

$S$ are plotted on the graph sheet as shown.

After observing the graph, We get,

The intersecting point of

$RS$ and

$PQ$ is

$T (2, 2)$ .

1690299_1794509_ans_ab56eab2f6ac4725b5c8f0ebe4d6654a.png

Find complement of the following:

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $57^o$ .

Then, the complement of $57^{0}=90^{0}-57^{0}=33^{0}$ .

Find the supplement of the following:

Supplement of

$87^{0}$ is

$180^{0}-87^{0}=93^{0}$ .

Read the following two statements which are taken as axioms:
(i) If two lines intersect each other, then the vertically opposite angles are not equal.
(ii) If a ray stands on a line, then the sum of two adjacent angles so formed is equal to $180^0$ .
Is this system of axioms consistent? Justify your answer.

The given system of axioms is not consistent because if a ray stands on a line and the sum of two adjacent angles so formed is equal to

$180^0$ . then for two lines which intersect each other, the vertically opposite angles becomes equal.

Find the complement of the following:

We know that,

if two angles are complementary, then their sum is

$90^o$ .

Given, the angle is

$25^o$ .

Then, the complement of

$25^{0}=90^{0}-25^{0}=65^{0}$ .

Find the complement of the following:

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $63^o$ .

Then, the complement of $63^{0}=90^{0}-63^{0}=27^{0}$ .

Find the supplement of the following:

Supplement of

$105^{0}$ is

$180^{0}-105^{0}=75^{0}$ .

The angles of a triangle are $x, y$ and $40^0$ the difference between the two angles $x$ and $y$ is $30^0$ Find $x$ and $y$ .

$x , y$ and

$40$ are the measure of interior angles of a triangle

$\therefore x +y + 40^0 =180^0$

$\Rightarrow x +y= 140^0 .....(i)$
The difference between

$x$ and

$y$ is

$30^0$ so

$x - y = 30^0 ...(ii)$

$x + y = 140^0 [from (i) ]$
______________________

$2x = 170^0$ [ adding

$(i)$ and

$(II)$ ]

$\Rightarrow x = \frac {170}{2} = 85^0$
Now

$x +y =140^0$ [ from (i) ]

$\Rightarrow 85^0 + y = 140^0 [ rs x = 85^0$ ]

$\Rightarrow y = 140^0 - 85^0$

$\Rightarrow y = 55^0$
and

$x = 85^0$

Find the supplement of the following:

Supplement of

$142^{0}$ is

$180^{0}-142^{0}=38^{0}$ .

Identify the following pairs of angles as complementary or supplementary: $112^{0},68^{0}$

$112^{0},68^{0}$ are supplementary.

$\because 112^o+68^o=180^o$

$72^{0},18^{0}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $72^o$ and $18^o$ .

Then, their sum $=72^o+18^o=90^o$ .

Therefore, they are complementary.

Find the value of $x$ in the above diagram:

In the given figure,

$2x+10+3x-10+40=180^{0}$

(Angles on one side of a straight line)

$\Rightarrow 5x+40^{0}=180^{0}$

$\Rightarrow 5x=180^{0}-40^{0}=140^{0}$

$\Rightarrow x=28^{0}$

Identify the following pairs of angles as complementary or supplementary: $55^{0},125^{0}$

$55^{0},125^{0}$ are Supplementary.

$\because 125^o+55^o=180^o$

Identify the following pairs of angles as complementary or supplementary: $137^{0},43^{0}$

$137^{0},43^{0}$ are supplementary.

$\because 137^o+43^o=180^o$

Two complementary angles are $(x+4)^{0}$ and $(2x-7)^{0}$ , find the value of $x$ .

Given,

$(x+4)^{0}$ and

$(2x-7)^{0}$ are complementary angles.

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then,

$x+4+2x-7=90^{0}$

$\Rightarrow 3x-3=90^{0}$

$\Rightarrow 3x=90^{0}+3=93^{0}$

$\Rightarrow x=31^{0}$ .

Find the value of $x$ in the above diagram:

In the given figure,

$3x+x+40^{0}=180^{0}$

(Angles on one side of a straight line)

$\Rightarrow 4x=180^{0}-40^{0}=140^{0}$

$\Rightarrow x=35^{0}$

$45^{0},45^{0}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $45^o$ and $45^o$ .

Then, their sum $=45^o+45^o=90^o$ .

Therefore, they are complementary.

Two supplementary angles are in ratio of $2:7$ , find the angles.

Two supplementary angles in the ratio

$2:7$ and the sum is

$180^{0}$

$\therefore$ First angle

$=\dfrac{180^{0}}{2+7}\times 2=\dfrac{180^{0}}{9}\times2=40^{0}$
and second angle

$=\dfrac{180^{0}}{9}\times7=140^{0}$ .

Identify whether the following pairs of angles is complementary:
$34^{0},56^{0}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are

$34^o$ and

$56^o$ .

Then, their sum

$=34^o+56^o=90^o$ .

Therefore, they are complementary.

Identify each of the given pair of angles as alternate interior angles, co-interior angles or corresponding angles or non of these in the given figure.
(i) $\angle 2,\angle 6$
(ii) $\angle 1,\angle 6$
(iii) $\angle 3,\angle 5$
(iv) $\angle 2,\angle 7$
(v) $\angle 3,\angle 6$
(vi) $\angle 4,\angle 8$

In the given figure,

(i)

$\angle 2,\angle 6$ - are corresponding angles.

(ii)

$\angle 1,\angle 6$ - none.

(iii)

$\angle 3,\angle 5$ - alternate interior angles.

(iv)

$\angle 2,\angle 7$ - none.

(v)

$\angle 3,\angle 6$ - co-interior angles.

(vi)

$\angle 4,\angle 8$ - corresponding angles.

If an angle is half of its complement, find the measure of angles.

We know that, if two angles are complementary, then their sum is

$90^o$ .

Let one angle =

$x^{0}$

Then, second angle =

$(\dfrac{x}{2})^o$

$\Rightarrow x+\dfrac{x}{2}=90^{0}$

$\Rightarrow \dfrac{3x}{2}=90^{0}$

$\Rightarrow 3x=90^{0}\times2=180^{0}$

$\Rightarrow x=60^{0}$ .

One angle

$=60^{0}$

and the other angle

$=\dfrac{x}{2}=\dfrac{60^o}{2}=30^{0}$ .

Therefore, the angles are

$30^{0}$ and

$60^{0}$ .

Two adjacent angles are in ratio $5:3$ and they together form an angle of $128^{0}$ , find these angles.

Sum of two adjacent angles =

$128^{0}$

and ratio between them =

$5:3$

Let first and =

$5x$

Then second angle =

$3x$

$5x+3x=128^{0}$

$\Rightarrow 8x=128^{0}$

$\Rightarrow x=16^{0}$

First angle =

$5x=5*16^{0}=80^{0}$

and second angle =

$3x=3*16^{0}=48^{0}$ .

Can two right angles be complementary?

We know that, if two angles are complementary, then their sum is

$90^o$ .

But given, the two angles are right angles, i.e.

$90^o, 90^o$ .

Then, their sum

$=90^o+90^o=180^o\ne90^o$ .

Therefore, the given statement is false.

Angles of a linear pair are as well as ..

Angles of a linear pair are as well as supplementary.

As their sum is

$180^o$

How many degrees are there in an angle which is one-fifth of its complement?

Let one angle =

$x$ .
But it is

$\dfrac{1}{5}$ th of its complementary angle,
Then, the complementary angle =

$5x$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

$\Rightarrow x+5x=90^{0}$

$\Rightarrow 6x=90^{0}$

$\Rightarrow x=15^{0}$ .

Required angle

$=15^{0}$ .

Adjacent angles have a common vertex, a common and no common _ points.

Adjacent angles have a common vertex, a common

$\underline{\text{arm}}$ , and no common

$\underline{\text{interior}}$ points.

Two angles forming a linear pair are .

Two angles forming a linear pair are supplementary as there sum is

$180^o$

Find the supplementary angle of the following angles:
$\frac{1}{2}$ of $90^{0}$

$\frac{1}{2}$ of

$90^{0}=45^{0}$
Supplementary angle of

$45^{0}=180^{0}-45^{0}=135^{0}$
Hence, supplementary angle of

$\frac{1}{2}$ of

$90^{0}$ is

$135^{0}$

Find the supplementary angle of the following angles:
$\frac{3}{7}$ of $280^{0}$

$\frac{3}{7}$ of

$280^{0}=120^{0}$
Supplementary of

$120^{0}$ is

$180^{0}-120^{0}=60^{0}$

If two angles are complementary, then the sum of their measures is ____.

We know that, if two angles are complementary, then their sum adds to

$90^o$ .

Therefore, if two angles are complementary, then the sum of their measures is

$90^o$ .

Supplement of an obtuse angle is .

Supplement of an obtuse angle is an acute angle.

If two adjacent angles are supplementary, then they form a ..

If two adjacent angles are supplementary, then they form a linear pair.

Write the complement angle of:
$\dfrac{1}{6}$ of $90^{o}$

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{1}{6}$ of $90^{o}$ is,

$=90^{o}-(\dfrac{1}{6} \times 90^{o})$

$=90^{o}-15^{o}$

$=75^{o}$ .

Therefore, the complement angle of $\dfrac{1}{6}$ of $90^{o}$ is, $75^{o}$ .

In the following figures, ROPE is parallelograms. Find the measures of angles $\mathrm{x}, \mathrm{y}$ and $\mathrm{z} .$ State the properties you use to find them.

From the figure, it is given that

$\angle \mathrm{POM}=80^{\circ}$ and

$\angle \mathrm{POE}=60^{\circ} .$
Then,

$\angle O P E=\angle P O M$ [because alternate angles are equal]

$\therefore y=80^{\circ}$

We know that, angles on the same straight line are equal to

$180^{\circ} .$

$\begin{array}{l}\angle \mathrm{ROE}+\angle \mathrm{EOP}+\mathrm{POM}=180^{\circ} \\\mathrm{x}+60^{\circ}+80^{\circ}=180^{\circ} \\\mathrm{x}+140^{\circ}=180^{\circ}\end{array}$

By solving we get,

$x=180^{\circ}-140^{\circ}$

$x=40^{\circ}$
Then,

$\angle \mathrm{ROE}=\angle \mathrm{OEP}$ [because alternate angles are equal]

$\begin{array}{l}x=z \\40^{\circ}=z\end{array}$
Therefore, value of

$x=40^{\circ}, y=80^{\circ}$ and

$z=40^{\circ}$

1708916_1810851_ans_56b584da34494deeaa87db9f06d27ff0.PNG

Write the complement angle of: $20^{o}+y^{o}$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then, the complement angle of

$20^{o}+y^{o}$ is,

$=90^{o}-(20^{o}+y^{o})$

$=90^{o}-20^{o}-y^{o}$

$=70^{o}-y^{o}$

$=(70-y)^{o}$ .

Hence, the required complement angle is

$(70-y)^{o}$ .

Write the complement angle of: $\dfrac{1}{2}$ of $60^{o}$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then, the complement angle of

$\dfrac{1}{2}$ of

$60^{o}$ is,

$=90^{o}-(\dfrac{1}{2} \times 60^{o})$

$=90^{o}-30^{o}$

$=60^{o}$ .

Therefore, the complement angle of

$\dfrac{1}{2}$ of

$60^{o}$ is

$60^{o}$ .

In the figure, given below name:
Two acute angles

In the given figure, the two acute abgles are

$\angle AOB$ and

$\angle AOD$

In the figure, given below name:
Three pairs of adjacent angles.

In the given figure, the three pairs of adjacent angles are as follows:

$\angle AOB$ and

$\angle BOC$

$\angle BOC$ and

$\angle COD$

$\angle COD$ and

$\angle DOA$

Write the complement angle of:
$\dfrac{2}{5}$ of $70^{o}$

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{2}{5}$ of $70^{o}$ is,

$=90^{o}-(\dfrac{2}{5} \times 70^{o})$

$=90^{o}-28^{o}$

$=62^{o}$ .

Therefore, the complement angle of $62^{o}$ .

Write the complement angle of: $\dfrac{1}{5}$ of $160^{o}$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{1}{5}$ of $160^{o}$ is,

$=90^{o}-(\dfrac{1}{5} \times 160^{o})$

$=90^{o}-32^{o}$

$=58^{o}$ .

Therefore, the complement angle of $\dfrac{1}{5}$ of $160^{o}$ is $58^{o}$ .

Write the complement angle of: $(x-10)^{o}$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then, the complement angle of

$(x-10)^{o}$ is,

$=90^{o}-(x-10)^{o}$

$=90^{o}-x^{o}+10^{o}$

$=100^{o}-x^{o}$

$=(100-x)^{o}$ .

Therefore, the complement of

$(x-10)^{o}$ is

$(100-x)^{o}$ .

In the figure, given below name:
Two obtuse angles

In the given figure, the two abtuse angles are

$\angle BOC$ and

$\angle COD$

In the given figure, $AOB$ is a straight line. Find the value of $x$ and also answer each of the following:
$\angle AOP =.....$

Given from figure.

$\angle AOP=x+30^o$

$\angle BOP=x-30^o$
Now,

$\angle AOP+\angle BOP=180^o$ (Since

$\angle AOB$ is a straight line)

$(x+30^o)+(x-30^o)=180^o$

$2x=180^o$
We get

$x=90^o$

$\angle AOP=x+30^o$

$=90^o+30^o$

$=120^o$

Two supplementary angles are in the ratio $7:11$ . Find the angles

Given
Two supplementary angles are in the ratio

$7:11$
Let the two supplementary angles be

$7x$ and

$11x$
Hence,

$7x+11x=180^{o}$

$18x=180^{o}$

$x=180^{o}/18$
We get,

$x=10^{o}$
So, two supplementary angles are

$7x=7\times 10^{o}$

$=70^{o}$

$11x=11\times 10^{o}$

$=110^{o}$
Therefore, two supplementary angles are

$70^{o}$ and

$110^{o}$

In the given figure, $PQR$ is a straight line. If:
$\angle PQS=110^o$ ; find $\angle RQS$

Given

$PQR$ is a straight line

$\angle PQS+\angle RQS=180^o$

$110^o +\angle RQS=180^o$
On further calculation, we get

$\angle RQS =180^o -110^o$

$\angle RQS =70^o$

In the given figure, $PQR$ is a straight line. If:
$\angle SQR=75^o$ ; find $\angle PQS$

Given

$PQR$ is a straight line
In the given figure,

$\angle PQS +\angle SQR =180^o$ [linear pair of angles]

$\angle PQS+75=180$
On further calculation, we get

$\angle PQS=180-75$

$\angle PQS=105^o$

In the figure given, name:
Two reflex angles.

We know, an angle which is more than

$180^{o}$ but less than

$360^{o}$ .

In the given figure, the two reflex angles are:
reflex

$\angle AOB$ and reflex

$\angle COB$ .

Other reflex angles are, reflex angle

$\angle AOD$ and reflex angle

$\angle COD$ .

The measures of two supplementary angles are $(3x+15)^{o}$ and $(2x+5)^{o}$ . Find $x$ .

Given

$(3x+15)^{o}$ and

$(2x+5)^{o}$ are two supplementary angles
We know that,
Sum of two supplementary angles

$=180^{o}$
Hence,

$(3x+15)^{o}+(2x+5)^{o}=180^{o}$

$3x+15+2x+5=180^{o}$

$5x-20^{o}=180^{o}$

$5x=180^{o}-20^{o}$

$5x=160^{o}$

$x=160^{o}/5$
We get,

$x=32^{o}$
Therefore, the value of

$x$ is

$32^{o}$

Find the angle:
that is equal to its supplement?

Let

$x$ be the angle equal to its supplement

Then,

$x+x=180^\circ\\\Rightarrow 2x=180\\\Rightarrow x=\dfrac{180}{2}\\\therefore\,x=90^\circ$

The angle equal to its supplement is

$90^{o}$

Two complementary angles are in the ratio $7:8$ . Find the angles

It is given that two complementary angles are in the ratio $7:8$ .

Consider $7x$ and $8 x$ as the angles.

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, $7x + 8x = 90^0$

$\implies$ $15x = 90^0$

$\implies$ $x = \dfrac{90}{15} = 6^0$ .

Here, the angles will be:

$7x=7\times 6^o=42^0$ and $8x= 8 \times 6 = 48^0$ .

Find the angle that is equal to its complement.

Let

$x$ be the angle equal to its complement.

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then,

$\Rightarrow x+x=90^\circ\\\Rightarrow 2x=90\\\Rightarrow x=\dfrac{90}{2}\\\Rightarrow\,x=45^\circ$ .

Therefore, the angle equal to its complement is

$45^{o}$ .

In the given diagram, $ABC$ is a straight line
If $x=53^o$ find $y$

From the figure,
Given that

$ABC$ is a straight line
Hence,

$\angle ABD+\angle DBC=180^o$ [Linear pair of angles]

$x+y=180^o$

$53^o+y=180^o$

$y=180^o-53^o$

$y=127^o$

In the given figure, $AOB$ is a straight line. Find the value of $x$ and also answer each of the following:
$\angle BOP =.....$

Given from figure.

$\angle AOP=x+30^o$

$\angle BOP=x-30^o$
Now,

$\angle AOP+\angle BOP=180^o$ (Since

$\angle AOB$ is a straight line)

$(x+30^o)+(x-30^o)=180^o$

$2x=180^o$
We get

$x=90^o$

$\angle BOP=x-30^o$

$=90^o-30^o$
We get

$=60^o$

Figure given below shows a pair of adjacent angles AOB and BOC. Find whether or not the exterior arms OA and OC are in the same straight line.

From given figure,

$\angle COB + \angle AOB = 88^\circ + 112^\circ$

We get

$= 200^\circ$ [which is not equal to

$180^\circ$ ]

Hence, the exterior arms OA and OC are not in the same straight line

Two straight lines AB and CD intersect each other at a point O and angle $AOC = 50^\circ,$ find:
$\angle BOC$

We know that,

The sum of adjacent angles is

$180^\circ$

So,

$\angle AOC + \angle BOC=180^\circ$

$\angle BOC = 180^\circ- 50^\circ$

Therefore,

$\angle BOC = 130^\circ$

Figure given below shows a pair of adjacent angles $AOB$ and $BOC$ . Find whether or not the exterior arms $OA$ and $OC$ are in the same straight line.

We know that,

The sum of adjacent angles

$AOB$ and

$COB$ is

$180^\circ$

Hence,

$\angle AOB + \angle COB = 180^\circ$

$(90^\circ - x) + (90^\circ + x) = 180^\circ$

$90^\circ - x + 90^\circ + x = 180^\circ$

We get,

$180^\circ = 180^\circ$

Thus the exterior arms

$OA$ and

$OC$ are in the same straight line.

Two straight lines AB and CD intersect each other at a point O and angle $AOC = 50^\circ,$ find:
$\angle AOD$

Given

$\angle AOC = 50^\circ$

We know that,

Vertically opposite angles are equal

So,

$\angle BOD = \angle AOC$

Therefore,

$\angle BOD = 50^\circ$

Now,

$\angle AOD + \angle BOD = 180^\circ$ (linear pair of angles)

$\implies \angle AOD + 50^\circ = 180^\circ$

$\implies \angle AOD = 180^\circ - 50^\circ=130^{\circ}$

We get,

$\angle AOD = 130^\circ$

For an angle $x^{o}$ , find:
the value of $x^{o}$ if its supplementary angle is three times its complementary angle.

A per the statement,
Supplementary angle

$=3$ (Complementary angle)

$180^{o}-x=3(90^{o}-x)$

$180^{o}-x=270^{o}-3x$

$-x+3x=270^{o}-180^{o}$

$2x=90^{o}$

$x=90^{o}/2$
We get,

$x=45^{o}$
Therefore, the value of

$x$ is

$45^{o}$

In the given figure, $AOB$ is a straight line. Find the value of $x$ and also answer each of the following:
Which angle is acute?

Given from figure.

$\angle AOP=x+30^o$

$\angle BOP=x-30^o$
Now,

$\angle AOP+\angle BOP=180^o$ (Since

$\angle AOB$ is a straight line)

$(x+30^o)+(x-30^o)=180^o$

$2x=180^o$
We get

$x=90^o$

$\angle AOP=x+30^o$

$=90^o+30^o$

$=120^o$

$\angle BOP=x-30^o$

$=90^o-30^o$

$=60^o$
The acute angle is

$\angle BOP$

In the given figure, $PQR$ is a straight line. Find $x$ . Then complete the following:
$\angle BQP =$ ________

Given

$PQR$ is a straight line

$\angle AQP =x+20^o$

$\angle AQB=2x+10^o$

$\angle BQP=x-10^o$
Since

$PQR$ is a straight line

$\angle AQP+\angle AQB +\angle BQR=180^o$

$(x+20^o)+(2x+10^o)+(x-10^o)=180^o$
We get,

$4x+20^o=180^o$

$4x=180^o-20^o$

$4x=160^o/4$
We get

$x=40^o$

$\angle AQB=2x+10^o$

$=2\times 40^o+10^o$

$=80^o+10^o$

$=90^o$
Similarly

$\angle AQP=x+20^o$

$\angle AQP=40^o+20^o$

$\angle AQP=60^o$

$\angle BQR=x-10^o$

$\angle BQR=40^o-10^o$

$\angle BQR=30^o$

We find,

$\angle BQP=\angle AQP+\angle AQB$

$=60^o +90^o$

$=150^o$

In the given figure, $PQR$ is a straight line. Find $x$ . Then complete the following:
$\angle AQB =$ ________

Given

$PQR$ is a straight line

$\angle AQP =x+20^o$

$\angle AQB=2x+10^o$

$\angle BQP=x-10^o$
Since

$PQR$ is a straight line

$\angle AQP+\angle AQB +\angle BQR=180^o$

$(x+20^o)+(2x+10^o)+(x-10^o)=180^o$
We get,

$4x+20^o=180^o$

$4x=180^o-20^o$

$4x=160^o/4$
We get

$x=40^o$

$\angle AQB=2x+10^o$

$=2\times 40^o+10^o$

$=80^o+10^o$

$=90^o$
Similarly

$\angle AQP=x+20^o$

$\angle AQP=40^o+20^o$

$\angle AQP=60^o$

$\angle BQR=x-10^o$

$\angle BQR=40^o-10^o$

$\angle BQR=30^o$

In the given figure, $AOB$ is a straight line. Find the value of $x$ and also answer each of the following:
Which angle is obtuse?

Given from figure.

$\angle AOP=x+30^o$

$\angle BOP=x-30^o$
Now,

$\angle AOP+\angle BOP=180^o$ (Since

$\angle AOB$ is a straight line)

$(x+30^o)+(x-30^o)=180^o$

$2x=180^o$
We get

$x=90^o$

$\angle AOP=x+30^o$

$=90^o+30^o$

$=120^o$

$\angle BOP=x-30^o$

$\angle 2$ and

$\angle 7$ = Alternate interior angles

A line segment AP stands at point P of a straight line BC such that $\angle APB = 5x - 40^\circ$ and $\angle APC = x + 10^\circ,$ find the value of $x$ and angle APB.

Given

A line segment AP stands at P and

$\angle APB = 5x - 40^\circ$

$\angle APC = x + 10^\circ$

(i) BPC is a straight line

$\angle APB + \angle APC = 180^\circ$

$5x - 40^\circ + x + 10^\circ = 180^\circ$

$6x - 30^\circ = 180^\circ$

$6x = 180^\circ + 30^\circ$

$6x = 210^\circ$

$x = 210^\circ / 6$

$x = 35^\circ$

(ii)

$\angle APB = 5x - 40^\circ$

$= 5 \times 35^\circ - 40^\circ$

$=175^\circ - 40^\circ$

$= 135^\circ$

1791094_1823242_ans_f1e359fbec5e4365a6e0ce37f49eca8b.png

Identify the pair of angles given below by referring the given figure as adjacent angles, vertically opposite angles, interior alternate angles, corresponding angles or exterior alternate angles.
$\angle 5$ and $\angle 7$

$\angle 5$ and

$\angle 7$ = Vertically opposite angles

In the given figure, find the following:
(i) $x$
(ii) $\angle AOB$
(iii) $\angle BOC$

We know that

$\angle AOB$ and

$\angle COB$ are linear pairs.
It can be written as

$\angle AOB + \angle COB = 180^{\circ}$
Substituting the values

$x +25^{\circ} + 3x +15^{\circ} = 180^{\circ}$
By further calculation

$4x +40^{\circ} = 180^{\circ}$
So we get

$4x = 180 - 40 = 140^{\circ}$
(i)

$x =\dfrac{ 140 }{ 4} = 35^{\circ}$
(ii)

$\angle AOB = x+25^{\circ}$
Substituting the value of x

$\angle AOB = 25^{\circ}+ 35^{\circ} = 60^{\circ}$
(iii)

$\angle BOC = 3x +15^{\circ}$
Substituting the value of x

$\angle BOC = ( 3 \times 35^{\circ}) + 15^{\circ}$

$\angle BOC = 120^{\circ}$

In which of the following figures, are $\angle AOB$ and $\angle AOC$ adjacent angles? Give, in each case, reason for your answer.

$\angle AOB$ and

$\angle AOC$ are adjacent angle , they have

$OA$ as their common arm.
(iv) From the figure

$OB$ is the common arm.

$\angle AOB$ and

$\angle AOC$ are not adjacent angles.

In which of the following figures, are $\angle AOB$ and $\angle AOC$ adjacent angles? Give, in each case, reason for your answer.

$\angle AOB$ and

$\angle AOC$ are adjacent angle , they have

$OA$ as their common arm.
(iii) From the figure

$OA$ is the common arm

$\angle AOB$ and

$\angle AOC$ are adjacent angles.

Find $y$ in the given figure.

Here

$AOC$ is a straight line
We can write it as

$\angle AOB + \angle BOD + \angle DOC = 180^0$
Substituting the values

$y +150 -x +x = 180$
By further calculation

$y + 150 = 180$
So we get

$y = 180 - 150 = 30^0$

Identify the pair of angles given below by referring the given figure as adjacent angles, vertically opposite angles, interior alternate angles, corresponding angles or exterior alternate angles.
$\angle 2$ and $\angle 8$

$\angle 2$ and

$\angle 8$ are pair of alternate exterior angles.

In which of the following figures, are $\angle AOB$ and $\angle AOC$ adjacent angles? Give, in each case, reason for your answer.

$\angle AOB$ and

$\angle AOC$ are adjacent angle , they have

$OA$ as their common arm.
(i) From the figure

$OB$ is the common arm,

$\angle AOB$ and

$\angle AOC$ are not adjacent angles.

In which of the following figures, are $\angle AOB$ and $\angle AOC$ adjacent angles? Give, in each case, reason for your answer.

$\angle AOB$ and

$\angle AOC$ are adjacent angle , they have

$OA$ as their common arm.
(ii) From the figure

$OC$ is the common arm

$\angle AOB$ and

$\angle AOC$ are not adjacent angle.

In the adjoining figure, if $b^o =a^o +c^o ,$ find $b$ .

It is given that

$b^0 = a^0 +c^0$
for a straight angle

$a^0 + b^0 + c^0 = 180^0$
Substituting the values

$b^0 +b^0 = 180^0$

$2b^0 = 180^0$

$b^0 = 180/2 = 90^0$

In the given figure, $AB$ is perpendicular to $BC$ at $B$ .
Find:
(i) the value of $x$ .
(ii) The complement of angle $x$ .

(i) From the figure,

$AB \perp BC$ at

$B$

Here

$\angle ABC = 90^o$ .
Substituting the values, we get,

$\implies$

$x +20^o +2x +1 ^o+ 7x - 11^o = 90^o$

$\implies$

$10x +10^o = 90^o$

$\implies$

$10x = 90^o - 10^o = 80^o$

$\implies$

$x = \dfrac{80}{10}$ .

$\implies$

$x = 8^o$ .

(ii) We know that, if two angles are complementary, then their sum is

$90^o$ .

Therefore, the complement of angle

$x = 90^o -x$

$= 90^o -8 = 82^o$ .

Write the complement of: $\dfrac{1}{2}$ of a right angle.

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{1}{2}$ of right angle $=\dfrac{1}{2}$ of $90^{o}$ is,

$=90^{o}-(\dfrac{1}{2} \times 90^{o})$

$=90^{o}-45^{o}$

$=45^{o}$ .

Therefore, the complement angle of $\dfrac{1}{2}$ of $90^{o}$ is $45^{o}$ .

Write the complement of the following angle:
$a^o$

The given angle is $a^{\circ}$ .

We know that,

If two angles are complementary, then their sum is $90^{\circ}$ .

So, the complement of $a^{\circ}=(90-a)^{\circ}$ .

Hence, the required answer is $(90-a)^{\circ}$ .

Write the complement of: $21^o \, 17'$ .

We know that, if two angles are complementary, then their sum is

$90^0$ .

Then, the complement angle of $21^0 \, 17'$ is,

$=90^{0}-$ $21^0 \, 17'$

$= 89^0 \, 60'$ $- 21^0 \, 17'$

$= 68^0 \, 43'$ .

Therefore, the complement angle of $21^0 \, 17'$ is $68^0 \, 43'$ .

Write the complement of: $(30-a)^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $(30-a)^o$ .

Then, the complement of $(30-a)^o$ is:

$=90^{0}-(30-a)^o=90^o-30^o+a^{0}$

$=60^{0}+a^o=(60+a)^o$ .

Write the complement of: $\dfrac{ 1}{3}$ of $180^o$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Then, the complement angle of $\dfrac{1}{3}$ of $180^{o}$ is,

$=90^{o}-(\dfrac{1}{3} \times 180^{o})$

$=90^{o}-60^{o}$

$=30^{o}$ .

Therefore, the complement angle of $\dfrac{1}{3}$ of $180^{o}$ is $30^{o}$ .

Write the complement of: $(x+5)^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $(x+5)^o$ .

Then, the complement of $(x+5)^o$ is:

$=90^{0}-(x+5)^{0}=90^o-x-5^o=85^o-x=(85-x)^o$ .

Write the complement of: $25^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $25^o$ .

Then, the complement of $25^{0}=90^{0}-25^{0}=65^{0}$ .

Write the complement of: $90^o$ .

We know that,

if two angles are complementary, then their sum is $90^o$ .

Given, the angle is $90^o$ .

Then, the complement of $90^{0}=90^{0}-90^{0}=0^{0}$ .

Write the supplement of:

$100^o$

The supplement of

$100^o$ is

$180^o -100^o$ which is equal to

$80^o .$

Write the supplement of:

$1/5$ of a right angle

The supplement of

$1 /5$ of a right angle

$= 180^0 -1/5$ of a right angle
we can write it as

$= 180^0 - 1/5 \times 90^0$
So we get

$= 180^0 -18^0$

$= 162^0$

Write the supplement of:

$(x+35)^o$

The Supplement of

$(x +35)^0 = 180^0 - (x +35)^0$
we can write it as

$= 180^o -x^o -35^o$

$37^0 \, 28'$ and

$52^o \, 33'$

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $37^0 \, 28'$ and $52^o \, 33'$ .

Then, their sum $37^0 \, 28'$ $+$ $52^o \, 33'$ $=$ $90^o \, 1' \ne 90^o$ .

Therefore, they are not complementary.

Write the supplement of:

$x^o$

The supplement of

$x^0 = 180^0 - x^0$

Are the following pairs of angles complementary?
$10^o$ and $80^o$ .

We know that, if two angles are complementary, then their sum is $90^o$ .

Given, the angles are $10^o$ and $80^o$ .

Then, their sum $=10^o+80^o=90^o$ .

Therefore, they are complementary.

Write the supplement of:

$(110-x-2y)^o$

The supplement of

$( 110 -x - 2y)^0 = 180^0 - ( 110 - x - 2y)^0$
we can write it as

$= 180 -110 +x +2y$

$=70^0 + x^0 +2y^0$

In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:
$\angle 4 \quad and \quad \angle 7$

$\angle 4 \quad and \quad \angle 7$ are interior alternate angles.

If $3x+18^o$ and $2x+25^o$ are supplementary, find the value of $x$ .

It is given that

$3x + 18060$ and

$2x +25^0$ are supplementary
we can write it as

$3x +18^0 + 2x +25^0 = 180^0$
By further calculation

$5x + 43^0 = 180^0$
So we get

$5x = 180 -43 = 137^0$

$x = 137/ 5 = 27.4^0$ or

$27^0 \ 24'$

Are the following pairs of angles supplementary?

$139^o \quad and \quad 39^o$

$139^o \quad and \quad 39^o$
No they are not supplementary angles as their sum is not equal to

$180^0$

$139^0 + 39^0 = 178^0$

Are the following pairs of angles supplementary?

$26^o 59' \quad and \quad 153^o 1'$

$26^o 59' \quad and \quad 153^o 1'$
yes they are supplementary angles as their sum

$= 26^0\ 59' + 153^0\ 1'= 180^0$

Are the following pairs of angles supplementary?

$2x^o +65^o \quad and \quad 115^o -2x^o$

$2x^0 +65^0$ and

$115^0 - 2x^0$
yes they are supplementary angles as their sum

$= 2x + 65 + 115 - 2x = 180^0$

Are the following pairs of angles complementary?
$(x+16)^o$ and $(74-x)^o$

Given,

$(x+16)^o$ and

$(74-x)^o$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

Then, their sum is

$= x + 16 + 74 - x = 90^0$ .

Hence, they are complementary.

In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:

$\angle 1 \quad and \quad \angle 4$

$\angle 1 \quad and \quad \angle 4$ are vertically opposite angles.

If two complementary angles are in the ratio $1:5$ , find them.

It is given that two complementary angles are in the ratio

$1 :5$ .

Consider

$x$ and

$5 x$ as the angles.

We know that, if two angles are complementary, then their sum is

$90^o$ .
Then,

$x + 5x = 90^0$

$\implies$

$6x = 90^0$

$\implies$

$x = \dfrac{90}{6} = 15^0$ .

Here, the angles will be

$15^0$ and

$15 \times 5 = 75^0$ .

In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:

$\angle 3 \quad and \quad \angle 7$

$\angle 3 \quad and \quad \angle 7$ are corresponding angles.

Are the following pairs of angles complementary?
$54^o$ and $\dfrac{2}{5}$ of a right angle.

We know that, if two angles are complementary, then their sum is

$90^o$ .

Given,

$54^o$ and

$\dfrac{2}{5}$ of a right angle.
We can write it as:

$=54^0$ and

$\dfrac25 \times 90^0$

$= 54^0$ and

$36^0$ .

Then, their sum

$= 54 +36 = 90^0$ .

Hence, they are complementary.

Find the value of 'x' in each of the following diagram.
$\angle COB = 90^o$

$\angle COD + \angle DOB = 90^o$

$4x + x = 90^o$

$5x = 90$

$x = \dfrac {90}{5}$

$x = 18^o$

Find the value of 'x' in each of the following diagram.
x + 2x = $180^o$

$x + 2x = 180^o$

$3x = 180^o$

$x = \dfrac {180^o}{3} = 60^o$

$x = 60^o$

Recognise type of angles in following figures.
Adjacent angles.

AOX is an acute angle and BOX is an obtuse angle.

Fill in the blanks with acute, right and straight :
An angle whose measure is greater than that of a right angle is _________

An angle whose measure is greater than that of a right angle is obtuse angle

Recognise the type of angles in the following figure.

Here,

$180^{\circ} < \angle AOB < 360^{\circ}$ .

We know, an angle whose measure is more than

$180^{\circ}$ but less than

$360^{\circ}$ , then it is called reflex angle.

Hence, the given, required angle

$AOB$ is a reflex angle.

1980015_1869879_ans_f2adcc61b954498586a7d7c401a90832.png

In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:

$\angle 11 \quad and \quad \angle 8$

$\angle 11 \quad and \quad \angle 8$ are allied or co-interior angles.

It is given that $\angle XYZ = 64^{\circ}$ and $XY$ is produced to point $P$ . Draw a figure from the given information. If ray $YQ$ bisects $\angle ZYP,$ find $\angle XYO$ and reflex $\angle QYP =?$

Given,

$\angle XYZ = 64^{\circ}$ and XY is produced upto

$P$ .

$\angle ZYP$ is bisected .
Also,

$YQ$ bisects

$\angle ZYP$

Let

$\angle PYQ = \angle ZYQ = x^{\circ}$
Then,

$\angle XYZ + \angle ZYQ + \angle QYP = 180^{\circ} ( \because \angle XYP$ is straight angle )

$\Rightarrow$

$64 + 2x = 180$

$\Rightarrow$

$2x = 180 - 64$

$\Rightarrow$

$2x = 116$

$\therefore x = \dfrac{116}{2} =58^{\circ}$ .

Then,

$\angle PYQ = \angle ZYQ = 58^{\circ}$ .
Reflex

$\angle QYP = \angle PYX +\angle XYZ + \angle ZYQ$

$=180^{\circ}+ 64^{\circ} + 58^{\circ}$

$=180 + 122=302^o$

$\therefore$ Reflex

$\angle QYP = 302^{\circ}$ .

1830024_1869675_ans_454424c0e6024df5957c704ed0c2f58b.PNG

In questions $1$ and $2,$ given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:

$\angle 6 \quad and \quad \angle 7$

$\angle 6 \quad and \quad \angle 7$ are adjacent angles.

Two angles which are not adjacent angles, but still supplementary.

1841716_1869874_ans_3527d473a95b4a39ba517dc84abe4280.png

Fill in the blanks with acute, right and straight :
An angle whose measure is greater than that of a right angle is ________.

Acute angle.

In figure, find the value of $x$ .

$\angle x + 140^{\circ} = 180^{\circ}$ (linear pair)

$\angle x=180^o-140^o$

$\Rightarrow x = 40^{\circ}$

If angles of magnitude (2x + 4) and (x - 1) from a linear pair. Find these angles.

According to question

$(2x + 4)+(x - 1) = 180^o$
(Linear pair axiom)

$\Rightarrow 3x + 3 = 180$

$\Rightarrow 3x = 177$

$x = \dfrac{177}{2} = 59$

$\therefore$ Angles are

$(2 \times 59 + 4)^o = 122^o$ and

$(59 - 1)^o = 58$

Find the magnitude of $\angle AOD$

$\angle AOC + \angle AOD = 180^o$
(Linear pair axiom)

$\Rightarrow 52^o + \angle AOD = 180^o$

$\Rightarrow \angle AOD = 180^o - 52^o$

$\Rightarrow \angle AOD = 128^o$

Name the adjacent supplementary angles of $\angle AOC$

The adjacent supplementary angles of

$\angle AOC$ are

$\angle AOD$ and

$\angle BOC$

Find the value of 'x' in each of the following diagram.
$\angle AOB = 180^o$

$\angle AOC + \angle BOC = 180$

$x-y+x+y =180$

$2x=180$

$x=\dfrac{180}{2}$

$x=90^o$

Which pair of angles are supplementary in the following diagram? Are they supplementary rays?

$\angle AOB$ and

$\angle COD$ are supplementary angles.

$\angle COB$ and

$\angle AOD$ are supplementary angles.
There are no supplementary rays in the diagram.

In the adjoining figure the sides $BA$ and $CA$ have been produced such that $BA = AD$ and $CA = AE$ . Prove that $DE | | BC$ (Hint: use the concept of alternative angles).

$\bigtriangleup AED$ and

$\bigtriangleup ABC$

$AE = AC$ (data )

$AD = AB$ (data )

$\angle EAD = angle BAC$
[ Vertically opposite angles ]

$\therefore \bigtriangleup AED = ABC$
[ SAS postulate ]

$\therefore \angle AED = \angle ACB$ [ Congruent triangle property ]
But these are alternative angles .

$\therefore DE | | BC$
Consequences of SAS postulate.

Write the pair of vertically opposite angles.

$(\angle AOC, \angle BOD)$ and

$(\angle AOD, \angle BOC)$ are vertically opposite pairs of angles.

Suppose two adjacent angles are supplementary. show that one of them is obtuse angle, then the other angle must be acute.

In the figure,

$\angle APQ + \angle BPQ = 180^o$ is an obtuse angle.
To prove

$\angle BPQ < 90^o$

Construction:
Draw a perpendicular SP at P
Proof:
In figure

$\angle APS + \angle SPQ + \angle QPS = 180$

$90 + \angle SPQ + \angle QPB = 180$

$\angle SPQ + \angle QPB = 180-90$

$\angle SPQ + \angle QPB = 90$
hence

$\angle SPQ < 90^o$ and

$\angle QPB<90^o$
hence

$\angle QPB$ is an acute angle
hence

$\angle APQ$ is an obtuse angle.

In the given figure, find $x$ and also find $\angle POS$

$\angle POS + \angle SOR + \angle ROQ = 180^{\circ}$ (straight angle)

$(x + 45^{\circ}) + x + (x + 30^{\circ} )= 180^{\circ}$

$\Rightarrow 3x + 75^{\circ} = 180^{ \circ}$

$\Rightarrow 3x = 105^{\circ}$

$\Rightarrow x = 35^{\circ}$

$\angle POS = (x + 45^{\circ} = (35^{\circ} + 45^{\circ}) = 80^{\circ}$

If the angles $(3x - 20)^{\circ}$ and $(2x - 40)^{\circ}$ are complementary angles. Find $x$ .

We know that the sum of the measure of an angle and its complement is equal to

$90^{\circ}$ .

Then,

$(3x - 20)^{\circ} + (2x - 40)^{\circ} = 90^{\circ}$

$\Rightarrow 5x - 60^{\circ} = 90^{\circ}$

$\Rightarrow 5x = 90^{\circ} +60^o$

$\Rightarrow 5x = 150^{\circ}$

$\Rightarrow x =30^{\circ}$ .

In the given figure, $POS$ is a line, find the measure of $\angle QOR.$

Here we are given that

$POS$ is a straight line.

$\angle POQ + \angle QOR + \angle ROS = 180^{\circ}$ (straight angle)

$\Rightarrow 60^{\circ} + 5x - 20^{\circ} + 40^{\circ} = 180^{\circ}$

$\Rightarrow 5x + 80^{\circ} = 180^{\circ}$

$\Rightarrow 5x = 100^{\circ}$

$\Rightarrow x = 20^{\circ}$

$\angle QOR = 5x -20^{\circ} = 5 \times 20^{\circ} - 20^{\circ} = 80^{\circ}$

An angle is $25^{\circ}$ less than its complement. What is it measure?

Let an angle be

$x$ .
Then, its complement

$= 90^{\circ} - x$ .

We know that, if two angles are complementary, then their sum is

$90^o$ .

According to the statement,

$x = 90^{\circ} - x - 25^{\circ}$

$\Rightarrow 2x = 65^{\circ}$

$\Rightarrow x=\dfrac{65}{2}$

$\Rightarrow x = 32\dfrac{1^{\circ}}{2}$ .

Hence, measurement of the required angle be

$32\dfrac{1^{\circ}}{2}$ .

In the figure, AO and BO are the bisectors of two adjacent angles A and B of a quad.ABCD. Prove that $2 \angle \mathrm{AOB}=\angle \mathrm{C}+\angle \mathrm{D}$

Given: In

$quad. ABCD$ the bisectors of

$\angle \mathrm{A}$ and

$\angle \mathrm{B}$ intersect at

$\mathrm{O}$ .
To prove:

$2 \angle \mathrm{AOB}=\angle \mathrm{C}+\angle \mathrm{D}$
Proof: In

$quad. ABCD$

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ} \ldots$ (i) (angle sum property of a quad.)

$\mathrm{Also}$ in

$\triangle \mathrm{AOB}$

$\angle \mathrm{OAB}+\angle \mathrm{OBA}+\angle \mathrm{AOB}=180^{\circ}($ by angle sum property of a

$\Delta)$

$\Rightarrow \dfrac{1}{2} \angle \mathrm{A}+\dfrac{1}{2} \angle \mathrm{B}+\angle \mathrm{AOB}=180^{\circ}$

$\Rightarrow \angle A+\angle B+2 \angle A O B=360^{\circ} \ldots($ ii

$)$
From (i) and (ii), we get

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=\angle \mathrm{A}+\angle \mathrm{B}+2 \angle \mathrm{AOB}$

$\Rightarrow 2 \angle \mathrm{AOB}=\angle \mathrm{C}+\angle \mathrm{D}$
Hence proved.
1882589_1880078_ans_821b568405a940c68fcf93249a97322d.png

In the figure, $\angle POM and \angle QOM$ from a linear pair. If $x - 2y = 30^{\circ},$ find $x$ and $y$ .

$\angle POM \;and \angle QOM$ from a linear pair.

$x + y = 180^{\circ}$ .....(i)
Also

$x - 2y =30^{\circ}$ (given)........(ii)
Solving (i) and (ii),

$3y=150$

$y=50$

From eq...(i)

$x=180-50=130$

we get

$x = 130^{\circ}$ and

$y = 50^{\circ}$

In the figure, $a ; b : c = 4 : 3 :$ If $AOB$ is a straight line, find the value of $a, b$ and $c$

We have,

$a ; b : c = 4 : 3 : 5.$

$\Rightarrow \angle a = 4x, \angle b = 3x \;and \angle c = 5x$
But

$\angle a + \angle b + \angle c = 180^{\circ}$

$\Rightarrow 4x + 3x +5x = 180^{\circ}$

$\Rightarrow 12x = 180^{\circ}$

$x = 15^{\circ}$
Angles are

$\angle a =4 \times 15^{\circ} = 60^{\circ},$

$\angle b= 3 \times 15^{\circ} =45^{\circ}$

and

$\angle c=5 \times 15^{\circ} = 75^{\circ}$

Evaluate x from figure

Given

$\angle ABD=\dfrac{3x}{2}$

$\angle ABC=x$

$\dfrac{3 x}{2}+x=180^{\circ}$ (Linear pair axiom)

$\Rightarrow 5 x=360^{\circ}$

$\Rightarrow x=72^{\circ}$

In the fig $AB=AC$ and $\angle ACD=120$ find $\angle A$ .

$\because AB=AC\Rightarrow \angle ABC=\angle ACB=x$ (say)

Let

$\angle BAC=y$

We know,

Exterior angles

$=$ sun of interior opposite angles

or,

$120^o =\angle ABC+\angle BAC$

or,

$120^o=x+y---(1)$

Again,

$\angle ACB+\angle ACD=180^o$

or,

$x+120^o=180^o$

$\therefore \ \boxed {x=60^o}$

From

$(1)$ ,

$60^o +y=120^o \Rightarrow \boxed {y=60^o} \therefore \boxed {\angle A=60^o}$

1385170_1127627_ans_009e2edc82f446f49e62b652357ea8e2.png

In given figure DE||BC and $AD:DB=5:4$ Find $\dfrac { Area(\triangle DEF) }{ Area(\triangle CFB) }$ .

$\triangle ABC$ , we have

DE||BC

$\Rightarrow$

$\angle ADE=\angle ABC$ and

$\angle AED=\angle ACB$ [Corresponding angles]

Thus, in triangles ADE and ABC, we have

$\angle A=\angle A$ [Common]

$\angle ADE=\angle ABC$

and,

$\angle AED=\angle ACB$

$\therefore$

$\triangle AED\sim \triangle ABC$ [By AAA similarity]

$\Rightarrow$

$\dfrac { AD }{ AB } =\dfrac { DE }{ BC }$

We have,

$\dfrac { AD }{ DB } =\dfrac { 5 }{ 4 }$

$\Rightarrow$

$\dfrac { DB }{ AD } =\dfrac { 4 }{ 5 }$

$\Rightarrow$

$\dfrac { DB }{ AD } +1=\dfrac { 4 }{ 5 } +1$

$\Rightarrow$

$\dfrac { DB+AD }{ AD } =\dfrac { 9 }{ 5 }$

$\Rightarrow$

$\dfrac { AB }{ AD } =\dfrac { 9 }{ 5 } \Rightarrow \dfrac { AD }{ AB } =\dfrac { 5 }{ 9 }$

$\therefore$

$\dfrac { DE }{ BC } =\dfrac { 5 }{ 9 }$

$\triangle DFE$ and

$\triangle CFB$ , we have

$\angle 1=\angle 3$ [Alternate interior angles]

$\angle 2=\angle 4$ [Vertically opposite angles]

Therefore, by AA-similarity criterion, we have

$\triangle DFE\sim \triangle CFB$

$\Rightarrow$

$\dfrac { Area(\triangle DFE) }{ Area(\triangle CFB) } =\dfrac { { DE }^{ 2 } }{ { BC }^{ 2 } }$

$\Rightarrow$

$\dfrac { Area(\triangle DFE) }{ Area(\triangle CFB) } ={ \left( \dfrac { 5 }{ 9 } \right) }^{ 2 }=\dfrac { 25 }{ 81 }$ .....[Using (i)]

1032069_1009451_ans_3a5df9b4ba314f97b40a55eb2e91784c.png

In the given figure,
If $\angle AOB=\angle BOC=\angle COD= 38^{\circ}$ ,
find reflex angle $\angle AOC$ and reflex angle $\angle AOD$ .

Given,

$\angle AOB=\angle BOC=\angle COD= 38^{\circ}$ .

Now,

$\angle AOC=\angle AOB+\angle BOC$

$=38^{\circ}+38^{\circ}=76^{\circ}$

$\implies \angle AOC={ 76^{ \circ } }$

Then, reflex angle

$\angle AOC={ 360^{ \circ } }-{ 76^{ \circ } }={ 284^{ \circ } }$ .

Now,

$\angle AOD=AOB+\angle BOC+\angle COD$

$=38^{\circ}+38^{\circ}+38^{\circ}={ 114^{ \circ } }$

$\implies \angle AOD={ 114^{ \circ } }$ .

Then, reflex angle

$\angle AOD={ 360^{ \circ } }-{ 114^{ \circ } }$

$={ 246^{ \circ} }$ .

In the diagram given below $\angle EDC$ . The tangent drawn to the circle at $C$ makes an angle of ${50}^{o}$ with $AB$ produced. Find the measure of $\angle ACB$ .

$\angle{AGE}={50}^{\circ}$

Consider

$AC$ diameter

$\triangle{ACG},\angle{ACG}={90}^{\circ}$

$\Rightarrow \angle{CAG}={40}^{\circ}$

$(sum~ of ~angle ~in ~triangle)$

$\triangle{ACB},\angle{CAB}={40}^{\circ}$

$\Rightarrow \angle{ACB}={50}^{\circ}$

In figure, $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ$ . $OS$ is another ray lying between rays $OP$ and $OR$ . Prove that $\angle ROS=\dfrac{1}{2}(\angle QOS -\angle POS)$ .

Given

$OR\bot PQ$ .

To Prove

$\angle ROS=\dfrac { 1 }{ 2 } \left( \angle QOS-\angle POS \right)$

Proof,

$\angle ROS=\angle ROP-\angle POS\quad \longrightarrow \left( i \right)$

also,

$\angle ROS=\angle QOS-\angle ROQ\quad \longrightarrow \left( ii \right)$

Now,

$\angle ROQ=\angle ROP$ [Both

${ 90 }^{ 0 }$ ]

$\therefore$ (ii)

$\Rightarrow$

$\angle ROS=\angle QOS-\angle ROP\quad \longrightarrow \left( iii \right)$

$(i)+(iii)$

$\Rightarrow$

$2\angle ROS=\angle QOS-\angle POS$

$\Rightarrow \angle ROS=\dfrac { 1 }{ 2 } \left( \angle QOS-\angle POS \right)$

Hence Proved.

1205824_1237808_ans_9d1ebe8b1ceb484bb3984cb58c2edc84.png

In fig. $6.57$ , $\angle Q>\angle R$ . $PA$ is the bisector of $\angle QPR$ and $PM\bot QR$ .Prove that $\angle APM=\dfrac{1}{2}\left(\angle Q-\angle R\right)$ .

Since PA is The bisector of

$\angle QPR$

$\angle QPR = \angle APR$

$\triangle PQM,\angle PQM+\angle PMQ+\angle QPM = 90^{\circ} \rightarrow (1)$

$\angle PQM+90+\angle QPM = 180^{\circ}$

$[\because PM\perp QE\Rightarrow \angle PMQ = 90^{\circ}]$

$\angle PQM = 90^{\circ}-\angle QPM \rightarrow (2)$

$\triangle PMR$

$\angle PMR +\angle PRM+\angle RPM = 180^{\circ}$ [by angle sum property of triangle]

$\Rightarrow 90^{\circ}+\angle PRM+\angle RPM = 180^{\circ}$

$[\because PM\perp QR\Rightarrow \angle PMR = 90^{\circ}]$

$\angle PRM = 90-\angle RPM \rightarrow (3)$

on subtracting (3) from (2) we get

$\angle Q-\angle R = (90-\angle QPM)-(90^{\circ}-\angle RPM)$

(where

$\angle PQM = \angle Q$ and

$\angle PRM = \angle R)$

$\angle Q-\angle R = \angle RPM-\angle QPM$

$\angle Q-\angle R = [\angle RPM+\angle APM]-[\angle QPA-\angle APM]\rightarrow (4)$

$\angle Q-\angle R = \angle QPA+\angle APM-\angle QPA+\angle APM$

$\angle Q-\angle R = 2\angle APM$

$\angle APM = \dfrac{1}{2}(\angle Q-\angle R)$ Hence proved

1186105_1164762_ans_b126eb253fa7403b9e08a7ef8565dd27.jpg

In figure, AB $||$ CD and AE $||$ CF, $\angle FCG=90^o$ and $\angle BAC=120^o$ . Find the values of x, y and z.

$||$ CD and AE

$||$ CF,

$\angle FCG=90^o$ and

$\angle BAC=120^o$

$line\ AB\ ||\ line\ CD$

$\therefore$

$\angle BAC$ and

$\angle ACD$ are supplementary angles.

Sum of supplementary angles is

$180^0$

$\therefore \angle BAC+\angle ACD=180^0$

$\therefore 120^0+x=180^0$

$\therefore x=60^0$

Also,

$\angle DCA+\angle ACF+\angle FCG=180^0$

$x+y+90^0=180^0$

$60^0+y+90^0=180^0$

$y=30^0$

$line\ AE\ ||\ line\ CF$

$\therefore$

$\angle EAC$ and

$\angle ACF$ are supplementary angles.

Sum of supplementary angles is

$180^0$

$\therefore \angle EAC+\angle ACF=180^0$

$\therefore z+y=180^0$

$\therefore z+30=180^0$

$\therefore z=150^0$

Construct the following angles using set-squares: $150^{\circ}$

Place the vertex of

$90^°$ of the set-square and make angle of

$90^°$ by drawing the rays

$BD$ and

$BC$ as shown in the above figure.

Then, place the vertex of

$60^°$ of the set square on the ray

$BD$ as shown in the figure and draw the ray

$BA$ .

The angle so formed is

$150^°$ .

Therefore,

$∠ABC = 150^°$

In figure, if array OC stands on line AB such that $\angle AOC=\angle COB$ , then show that $\angle AOC={ 90 }^{ \circ }$

1977329_1495556_ans_d394b1dd0c9b4087a03220f57f97700a.jpg

In Fig., $\angle AOF$ and $\angle FOG$ form a linear pair. If $\angle EOB = \angle FOC = 90^{\circ}$ and $\angle DOC =\angle AOB = 30^{\circ}.$
Name three pairs of adjacenet angles.

Adjacent Angles are those angles that share a common side and a common point (vertex) and they shouldn’t overlap

Hence

$\angle BOC, \angle COD; \angle COD, \angle DOE; \angle DOE, \angle EOF$ are pairs of adjacent angles.

In figure, find out $\angle PXR$ , if PQ $||$ RS.

1615748_1677584_ans_af63129aad7b420caa26ab39e7a858e4.jpg

In the figure, $AOB$ is a straight line. Find the value of $x$ . Here, find $\angle AOC,\angle COD$ and $\angle BOD$

From the figure, we know that

$\angle BOD$ and

$\angle AOD$ are linear pair of angles

So we get

$\angle BOD+\angle AOD={180}^{o}$

It can also be written as

$\angle BOD+\angle COD+\angle COA={180}^{o}$

By substituting the values

${ x }^{ o }+{ \left( 2x-19 \right) }^{ o }+{ \left( 3x+7 \right) }^{ o }={ 180 }^{ o }$

$\Rightarrow x+2x-{ 19 }^{ o } +3x+{ 7 }^{ o }={ 180 }^{ o }$

$\Rightarrow 6x-{ 12 }^{ o }={ 180 }^{ o }$

$\Rightarrow 6x={ 192 }^{ o }\Rightarrow x={ 32 }^{ o }$

By substituting the value of

$x$ we get

$\angle AOC={ \left( 3x+7 \right) }^{ o }=3({32}^{o})+{7}^{o}={103}^{o}$

$\angle COD={ \left( 2x-19 \right) }^{ o }=2({32}^{o})-{19}^{o}={45}^{o}$

$\angle BOD={x}^{o}={32}^{o}$

In the given figure, $p^\circ =q^\circ =r^\circ ,$ find each angle.

We know that

$p^0 + q^0 + r^0 = 180^0$ is the straight angle.
its is given that

$p^0 = q^0 = r^0$
We can write it as

$p^0 +p^0 +p^0 = 180^0$

$3p = 180$

$p = 180/ 3 = 60^0$
therefore

$, p^0 = q^0 = r^0 = 60^0$

In each of the above figure write, if any, $(i)$ each pair of vertically opposite angles, and $(ii)$ each linear pair.

1900203_1790538_ans_ae46f21682c4420aa0371c32775d97ff.jpg

In the figure, $AOB$ is a straight line, Find the value of $x$ . Hence, find $\angle AOC$ and $\angle BOD$ .

From the figure we know that

$\angle AOB$ is a straight line

So we get

$\angle AOB={180}^{o}$

It can also be written as

$\angle AOC+\angle COD+\angle BOD={180}^{o}$

By substituting the values

${(3x-7)}^{o}+{55}^{o}+{(x+20)}^{o}={180}^{o}$

$3x-{7}^{o}+{55}^{o}+x+{20}^{o}={180}^{o}$

On further calculate

$4x+{180}^{o}={180}^{o}$

$4x={112}^{o}$

$x={28}^{o}$

By substituting the value of

$x$ we get

$\angle AOC={(3x-7)}^{o}=3({28}^{o})-{7}^{o}$

$={84}^{o}-{7}^{o}={77}^{o}$

$\angle BOD={(x+20)}^{o}={(28+20)}^{o}={48}^{o}$

In the given figure, $PQR$ is a straight line. Find $x$ . Then complete the following:
$\angle AQR =$ ________

Given

$PQR$ is a straight line

$\angle AQP =x+20^o$

$\angle AQB=2x+10^o$

$\angle BQP=x-10^o$
Since

$PQR$ is a straight line

$\angle AQP+\angle AQB +\angle BQR=180^o$

$(x+20^o)+(2x+10^o)+(x-10^o)=180^o$
We get,

$4x+20^o=180^o$

$4x=180^o-20^o$

$4x=160^o/4$
We get

$x=40^o$

$\angle AQB=2x+10^o$

$=2\times 40^o+10^o$

$=80^o+10^o$

$=90^o$
Similarly

$\angle AQP=x+20^o$

$\angle AQP=40^o+20^o$

$\angle AQP=60^o$

$\angle BQR=x-10^o$

$\angle BQR=40^o-10^o$

$\angle BQR=30^o$

We find,

$\angle AQR=\angle AQB+\angle BQR$

$=90^o +30^o$

$=120^o$

	$Angle - 1$	$Angle - 2$
(1)	$50^{\circ}$	_________
(2)	____________	$63^{\circ}$
(3)	$47^{\circ}$	_________
(4)	$56^{\circ}$	_________
(5)	__________	$12^{\circ}$
(6)	$67^{\circ}$	_________

Lines And Angles - Class 9 Maths - Extra Questions

Draw a line segment ABAB of 44 cm in length. Draw a line perpendicular to ABAB through AA and BB, respectively. Are these lines parallel?If true write 11 and if false then write 00

Can two acute angles be complement to each other always?If true, then enter 11 and if false, then enter 00.

If true then enter 11 and if false then enter 00Can two obtuse angles be adjacent angles?

If true then enter 11 and if false then enter 00Can two right angles be supplementary?

If true then enter 11 and if false then enter 00Can two adjacent angles be supplementary?

Measure of an angle is given below. Find the measure of it's complementary angle:58^o58^o.

If true then enter 11 and if false then enter 00Can two acute angles form a linear pair?

Measures of some angles are given below. Find the measures of their supplementary angles.108^o^o

Measures of some angles are given below. Find the measure of their complementary angle:5656^o^o

Measures of some angles are given below. Find the measures of their complementary angles:16^o^o.

Measures of some angles are given below. Find the measures of their complementary angles:6868^o^o

Find the measure of the supplementary angle pair of 60^o^o. (in degrees)

Measures of some angle is given below. Find the measures of its complementary angle:If the angle is \displaystyle \frac{2}{3}\displaystyle \frac{2}{3}rd of a right angle.

Find the angle which is equal to its complement.(in ^\circ^\circ)

Find the measures of the third angle in degrees \\ 92^{\circ}, 27^{\circ},----\\ 92^{\circ}, 27^{\circ},----

Side BC of a \Delta ABC\Delta ABC is produced in both the directions. Prove that the sum of the two exterior angles so formed is greater than 180^o180^o.

When do you say two angles are adjacent?

Observe the following figures and identify the type of angles in them

Identify the following given angles as acute , right or obtuse.

In each of the following figure AB || CD. Find the value of x in each case.

Find the supplement of the following angles.58^{\circ}58^{\circ}

In the given figure if AB || DE, \angle BAC = 35^o\angle BAC = 35^o and \angle CDE = 53^o\angle CDE = 53^o, find \angle DCE\angle DCE.

To prove: \angle 1\angle 1 is supplementary to \angle 8\angle 8. Write reasons for the statement.

Statement Reason\angle 1 + \angle 8 = 180^o\angle 1 + \angle 8 = 180^o .......................

Identify the angles which are more than right angles and which are less than right angle?

In each of the following figure AB || CD. Find the value of x in each case.

Find the complementary angle of 36^{\circ}36^{\circ}.

Can two adjacent angles be complementary ? Draw figure .

Can two adjacent angles be supplementary ? Draw figure .

Write the measure of supplementary angle of the given angles:47^{\circ}47^{\circ}.

Which of the following pairs of angles are supplementary ?

Find the value of unknown 'xx' in the following triangle:

Fill up the table on finding complementary angle of given angle:Angle - 1Angle - 1Angle - 2Angle - 2(1)50^{\circ}50^{\circ}_________(2)____________63^{\circ}63^{\circ}(3)47^{\circ}47^{\circ}_________(4)56^{\circ}56^{\circ}_________(5)__________12^{\circ}12^{\circ}(6)67^{\circ}67^{\circ}_________

Write the measure of supplementary angle of the given angles:75^{\circ}75^{\circ}.

Measure of one angle of linear pair is given. Find the measure of another angle:111^{\circ}111^{\circ}.

Write the measure of the supplementary angle of the given angle:112^{\circ}112^{\circ}.

Write the measure of supplementary angle of the given angles:100^{\circ}100^{\circ}.

Write the measure of supplementary angle of the given angles:81^{\circ}81^{\circ}.

Write the measure of supplementary angle of the given angles:109^{\circ}109^{\circ}.

Write the measure of supplementary angle of the given angles:90^{\circ}90^{\circ}.

Write the measure of supplementary angle of the given angles:132^{\circ}132^{\circ}.

Write the measure of supplementary angle of the given angles:60^{\circ}60^{\circ}.

Measure of one angle of linear pair is given. Find the measure of another angle:20^{\circ}20^{\circ}.

Measure of one angle of linear pair is given. Find the measure of another angle:130^{\circ}130^{\circ}.

m\angle XVW = 120^{\circ}m\angle XVW = 120^{\circ}, then(i) m\angle WVY =m\angle WVY = ______ and(ii) m\angle XVZ =m\angle XVZ = ______.

How may pairs are formed of angles of linear pair?

Find the angles x and y in the figure. AO=OBAO=OB

Find x, y, z.

In the figures given, state whether the lines ll and mm are parallel or not.

Show that the angles of an equilateral triangle are 60^o60^o each.

Can two obtuse angles be supplementary?

Find the supplement of 105.105.

If PQ\bot PS,\ PQ\parallel SR,\ \angle SOR = 28^{o}PQ\bot PS,\ PQ\parallel SR,\ \angle SOR = 28^{o} and \angle QRT = 65^{o}\angle QRT = 65^{o}, then find xx & yy.

Find the supplementary of 37^028' 37^028' and 52^0 33'52^0 33'.

In Fig. if AB || DE, \angle BAC = 35^oAB || DE, \angle BAC = 35^o and angle CDE = 53^oangle CDE = 53^o, find \angle DCE\angle DCE.

If the sum of the measures of two angles is {90^\circ}{90^\circ}, then the angles are called _____ angles.

Find the angles, when:(i)The angles are complementary and the smaller is 40^ \circ40^ \circ less than the larger. (ii)The angles are complementary and the larger is four times the smaller.

In the figure \angle ABC={80}^{o}\angle ABC={80}^{o}. Find \angle DGC\angle DGC and \angle DEF\angle DEF, given AB||DEAB||DE and BC||EFBC||EF.

Find a.a.

Name the pairs of adjacent angles in the figures.

Manasa says, "Each angle in any pair of complementary angles is always acute". Do you agree? Give reason.

AE||BD, AC||EDAE||BD, AC||ED & AB=ACAB=AC find a,b,ca,b,c

Find \angle APB\angle APB, if AE || BRAE || BR.

In the figure,\Delta PQR\Delta PQR is a right angled at Q, \overline {ML} ||\overline {RQ} \overline {ML} ||\overline {RQ} and \angle LMR = {130^ \circ }\angle LMR = {130^ \circ }. Find \angle LPM\angle LPM and \angle PML \angle PML

Find the measure of an angle which measures same as its complement.

AD is an altitude of an isosceles triangle ABC in which AB=AC show that, (i) AD bisects BC (ii) AD bisects \angle A\angle A.

If A,B,CA,B,C are the angles of a triangle, show that\sin{B}\cos{(C+A)}+\cos{B}\sin{(C+A)}=0\sin{B}\cos{(C+A)}+\cos{B}\sin{(C+A)}=0

In \triangle\ PQR,\angle\ Q=40^{o},\angle\ R=75^{o}\triangle\ PQR,\angle\ Q=40^{o},\angle\ R=75^{o}, then find the shortest and the longest side of the triangle

In the adjoining figure, name the following pairs of angles.Equal supplementary angles

Explain:Can two acute angles be supplementary

Can two right angles be supplementary?

Two complementary angles are in the ratio of 2:12:1. Find the measure of the angles.

In the adjoining figure, name the following pairs of angles.Unequal supplementary angles

In the following figure, is \angle 1\angle 1 adjacent to \angle 2\angle 2? Given reasons.

Find the angles of a triangle that are in ratio 2 : 2 : 52 : 2 : 5.

Find the value of xx

Two adjacent angle of ||gm||gm are 2x-15^o2x-15^o and 7x-75^o7x-75^o.Find all angles of ||gm||gm ?

Find each one of lettered angles giving reason

Find x in the following figure (AB || CD)(AB || CD) :

Draw a line segment $AB$ of $4$ cm in length. Draw a line perpendicular to $AB$ through $A$ and $B$ , respectively. Are these lines parallel?
If true write $1$ and if false then write $0$

Can two acute angles be complement to each other always?
If true, then enter $1$ and if false, then enter $0$ .

If true then enter $1$ and if false then enter $0$
Can two obtuse angles be adjacent angles?

If true then enter $1$ and if false then enter $0$
Can two right angles be supplementary?

If true then enter $1$ and if false then enter $0$
Can two adjacent angles be supplementary?

Measure of an angle is given below. Find the measure of it's complementary angle:
$58^o$ .

If true then enter $1$ and if false then enter $0$
Can two acute angles form a linear pair?

Measures of some angles are given below. Find the measures of their supplementary angles.

108 $^o$

Measures of some angles are given below. Find the measure of their complementary angle:
$56$ $^o$

Measures of some angles are given below. Find the measures of their complementary angles:
16 $^o$ .

Measures of some angles are given below. Find the measures of their complementary angles:
$68$ $^o$

Find the measure of the supplementary angle pair of 60 $^o$ . (in degrees)

Measures of some angle is given below. Find the measures of its complementary angle:
If the angle is $\displaystyle \frac{2}{3}$ rd of a right angle.

equal to its complement.(in $^\circ$ )

degrees $\\ 92^{\circ}, 27^{\circ},----$

Side BC of a $\Delta ABC$ is produced in both the directions. Prove that the sum of the two exterior angles so formed is greater than $180^o$ .

Find the supplement of the following angles.
$58^{\circ}$

In the given figure if AB || DE, $\angle BAC = 35^o$ and $\angle CDE = 53^o$ , find $\angle DCE$ .

To prove: $\angle 1$ is to $\angle 8$ . Write reasons for the statement.

Statement Reason
$\angle 1 + \angle 8 = 180^o$ .......................

Find the complementary angle of $36^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$47^{\circ}$ .

Find the value of unknown ' $x$ ' in the following triangle:

Fill up the table on finding complementary angle of given angle:
$Angle - 1$ $Angle - 2$
(1) $50^{\circ}$ _
(2) $63^{\circ}$
(3) $47^{\circ}$ _
(4) $56^{\circ}$ _
(5) $12^{\circ}$
(6) $67^{\circ}$ ___

Write the measure of supplementary angle of the given angles:
$75^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$111^{\circ}$ .

Write the measure of the supplementary angle of the given angle:
$112^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$100^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$81^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$109^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$90^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$132^{\circ}$ .

Write the measure of supplementary angle of the given angles:
$60^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$20^{\circ}$ .

Measure of one angle of linear pair is given. Find the measure of another angle:
$130^{\circ}$ .

$m\angle XVW = 120^{\circ}$ , then
(i) $m\angle WVY =$ and
(ii) $m\angle XVZ =$ .

Find the angles x and y in the figure. $AO=OB$

In the figures given, state whether the lines $l$ and $m$ are parallel or not.

Show that the angles of an equilateral triangle are $60^o$ each.

Find the supplement of $105.$

If $PQ\bot PS,\ PQ\parallel SR,\ \angle SOR = 28^{o}$ and $\angle QRT = 65^{o}$ , then find $x$ & $y$ .

Find the supplementary of $37^028'$ and $52^0 33'$ .

In Fig. if $AB || DE, \angle BAC = 35^o$ and $angle CDE = 53^o$ , find $\angle DCE$ .

If the sum of the measures of two angles is ${90^\circ}$ , then the angles are called _____ angles.

Find the angles, when:
(i)The angles are complementary and the smaller is $40^ \circ$ less than the larger.
(ii)The angles are complementary and the larger is four times the smaller.

In the figure $\angle ABC={80}^{o}$ . Find $\angle DGC$ and $\angle DEF$ , given $AB||DE$ and $BC||EF$ .

Find $a.$

$AE||BD, AC||ED$ & $AB=AC$ find $a,b,c$

Find $\angle APB$ , if $AE || BR$ .

In the figure, $\Delta PQR$ is a right angled at Q, $\overline {ML} ||\overline {RQ}$ and $\angle LMR = {130^ \circ }$ . Find $\angle LPM$ and $\angle PML$

AD is an altitude of an isosceles triangle ABC in which AB=AC show that, (i) AD bisects BC (ii) AD bisects $\angle A$ .

If $A,B,C$ are the angles of a triangle, show that
$\sin{B}\cos{(C+A)}+\cos{B}\sin{(C+A)}=0$

In $\triangle\ PQR,\angle\ Q=40^{o},\angle\ R=75^{o}$ , then find the shortest and the longest side of the triangle

In the adjoining figure, name the following pairs of angles.
Equal supplementary angles

Explain:
Can two acute angles be supplementary

Two complementary angles are in the ratio of $2:1$ . Find the measure of the angles.

In the adjoining figure, name the following pairs of angles.
Unequal supplementary angles

In the following figure, is $\angle 1$ adjacent to $\angle 2$ ? Given reasons.

Find the angles of a triangle that are in ratio $2 : 2 : 5$ .

Find the value of $x$

Two adjacent angle of $||gm$ are $2x-15^o$ and $7x-75^o$ .Find all angles of $||gm$ ?

Find x in the following figure $(AB || CD)$ :

Find the complements of the following angles:-
a) ${45^ \circ }$
b) ${8^ \circ }$
c) ${20^ \circ }$
d) ${25^ \circ }$
e) ${78^ \circ }$

Calculate the value of $x :$

Find the value of $x$ ?

In the figure $l\parallel m$ and $p$ is a transversal, find the value of $x$ , giving necessary reasons.

In figure given alongside $AB$ and $CD$ are intersecting lines. Name the following pair of angles.
Adjacent angles that do not form a linear pair and each less than $180^{o}$ .

From the adjoining figure find $x$ , i $AOB$ is a straight line. Hence complete the following
Which is an obtuse angle?

The supplementary angle of $40^{o}$ is

From the adjoining figure find $x$ , i $AOB$ is a straight line. Hence complete the following
Which is an acute angle?

Fill in the blank.
The sum of the measure of supplementary angle is ______

Find the supplement of the angle whose measure is ${60}^{\circ}$

In figure given alongside $AB$ and $CD$ are intersecting lines. Name the following pair of angles.
Equal supplementary angles

In Fig. If $x+y=w+z$ then prove that $A O B$ is a line.

If the ratio between two complementary angles $2:3$ , then find the angles.

Find the supplementary angle of $105^{o}$ .

Find the supplementary angle of $115^{o}$ .

Find the supplementary angle of $87^{\circ}$

Find the supplementary angle of $40^o$ .

For an angle ${ x }^{ 2 }$ , find: The complementary angle.