Class 7 Maths - Lines And Angles

In Fig., lines AB and CD are parallel and P is any point as shown in the figure. Show that $\angle ABP+\angle CDP= \angle DPB$ .

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In figure, write another name for $\angle 2$ .

$\angle 2$ is nothing but

$\angle BOC.$

Find $x$ and $y$ .

According to question,

By alternate angle,

$x=y$

In a triangle,

The sum of all angles of a triangle is

$\pi.$

Therefore,

$x+x+x=\pi$

$=>x=\dfrac{\pi}{3}$

Also,

$y=\dfrac{\pi}{3}$

In the adjoining figure, name the following pairs of angles.
Obstuse vertically opposite angles

We have,

In the given figure,

$\angle\ AOB=\angle\ COD$
Obtuse vertically opposite angle.

Hence, this is the answer.

In the given figure three lines $\overleftrightarrow{AB}, \overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$ intersecting at $O$ . Find the values of $x, y$ and $z$ it is being given that $x : y : z = 2 : 3 : 5$

$x:y:z=2:3:5$

let the angles be

$2u,3u$ and

$5u$

$\angle BOF=\angle AOE={ x }^{ \circ }={2u}$ (Alternete angles)

$\angle EOD=\angle COF={ y }^{ \circ }={3u}$ (Alternete angles)

$\angle AOC=\angle DOB={ z }^{ \circ }={5u}$ (Alternete angles)

$\angle EOD+\angle DOB+\angle BOF+\angle FOC+\angle COA+\angle AOE+\angle ={360}^{\circ}$

$3u+5u+2u+3u+5u+2u={360}^{\circ}$

$20u={360}^{\circ}$

$u=\dfrac{{360}^{\circ}}{20}={18}^{\circ}$

$x=2\times{18}^{\circ}={36}^{\circ}$

$y=3\times{18}^{\circ}={54}^{\circ}$

$z=5\times{18}^{\circ}={90}^{\circ}$

Two complementary angles are in the ratio $4 : 5$ . Find the angles.

Let the two angles be

$4x$ and

$5x$

The sum of two complementary angles is

${ 90 }^{ \circ }$

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

$\Rightarrow 9x={ 90 }^{ \circ }$

$\Rightarrow x=\dfrac { { 90 }^{ \circ } }{ 9 } ={ 10 }^{ \circ }$

So the angles are

$4\times{ 10 }^{ \circ }={ 40 }^{ \circ }$ and

$5\times { 10 }^{ \circ }={ 50 }^{ \circ }$

In the figure, find the values of $x$ and $y$ and then show that $AB\parallel CD$ .

$x + 50^{\circ} = 180^{\circ}$ (Linear pair)

$\Rightarrow x = 130^{\circ}$

Also,

$y = 130^{\circ}$ (Vertically opposite)

Now,

$x = y = 130^{\circ}$ (Alternate interior angles)

Alternate interior angles are equal.

$\therefore AB \parallel CD.$

Vertically opposite angle of $\angle XVW$ is _______.

$\angle YVZ$ .

Write all pairs of vertically opposite angles.

$\angle XMQ$ and

$\angle PMY, \angle XMP$ and

$\angle YMQ$ .

Vertically opposite angle of $\angle XVZ$ is __________.

Vertically opposite angle of given angle is

$\angle WVY$ .

Find x:

We can see that,

$(2+3x)^o$ and

$62^o$ are Vertically opposite Angle

So,

$(2+3x)^o=62^o\Rightarrow 3x=60^o\Rightarrow x=20^o$

How many pair of vertically opposite angles are formed?

There are two pairs of vertically opposite angle in the provided diagram as:

$\angle ACY = \angle XCB,\angle ACX = \angle BCY$

Write the pair of vertically opposite angles.

$\angle ACY$ and

$\angle XCB, \angle ACX$ and

$\angle BCY$ are two pairs of vertically opposite angles.

In the given figure , the arms of two angles are parallel. if $\angle ABC = {70^ \circ }$ , then find
(i) $\angle DGC$
(ii) $\angle DEF$

Given that:

Sides of the two angles are parallel so

$AB\parallel DGE$ and

$line\ BGC\parallel EF$

$i)$ If

$AB\parallel DGE$ and

$BC$ is transversal line, then

$\angle DGC=\angle ABC=70^\circ$ (Corresponding angle)

$ii)$ If

$line\ BGC\parallel EF$ and

$DE$ is transversal line, then

$\angle DEF=\angle DGC=70^\circ$ (Corresponding angle)

Fill in the blanks:
If two line intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are_______.

If two line intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles.

In the adjoining figure, $p\parallel q$ . Find the unknown angles.

$p \parallel q$ then

$125^{\circ}+\angle e=180^{\circ}$ (linear pair)

$\angle e=55^{\circ}$

$\angle f=55^{\circ}$ (vrtically opposite angles)

$\angle a=\angle e$

$(p \, parallel \ to\, q)$

$\angle a=55^{\circ}$

$\angle d=125^{\circ}$

$(p \, parallel \, to\, q)$

$\angle c =\angle f=55^{\circ}$

$\angle b+\angle c=180^{\circ}$ (linear pair).

$\angle b=125^{\circ}$

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In the given figure, if $\angle A O C + \angle B O D = 266 ^ { \circ }$ find all the four angles.

Given

$\angle AOC+\angle BOD=266°$

$\angle AOC=\angle BOD$ (opposite angles are same)

$\angle AOC+\angle BOD=266°$

$\Rightarrow 2\angle AOC=266°$

$\Rightarrow \angle AOC=133°$

$\angle AOC+\angle AOD=180°$ (complementary angles)

$133°+\angle AOD=180°$

$\Rightarrow \angle AOD=180°-133°$

$\Rightarrow \angle AOD=47°$

$\angle BOC=47°$ (opposite angle of

$\angle AOD$ )

$\therefore\ \angle AOC=\angle BOD=266°,\angle AOD=\angle BOC=47°$

In the given figure, find $x$ .

REF.Image

Opposite angle are equal

$\therefore x=30^{\circ}$

Now,

$30+30+y=180^{\circ}$

$y=120^{\circ}$

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Corresponding angles are on the _________ side of the transversal.

Corresponding angles are on the

$\text{same}$ side of the transversal.

The sum of two vertically opposite angles is $166^\circ.$ Find each of the angles.

As we know, the vertically opposite angles are equal.

Let

$y$ be the angle.

As the sum of angles is given as

$166^\circ,$

$y+y=166^o$

$\Rightarrow 2y=166^\circ$

$\Rightarrow y=\dfrac{166^\circ}{2}$

$\Rightarrow y=83^\circ$

Therefore, both angles are of

$83^\circ.$

A road crosses a railway line at an angle of $30^o$ as shown in Fig.
Find the values of $a,\ b$ and $c$ .

In given figure,

$l$ parallel to

$m$ and

$q$ is a transversal.

$\therefore \angle 1 = 30^o$ ......

$(1)$ [corresponding angles]
Here,

$m$ is a straight line.

$\therefore \angle b +\angle 1=180^o$ [Linear pair]

$\Rightarrow \angle b=180^o - 30^o =150^o$ by using equation

$1$
Similarly,

$p$ parallel to

$q$ and

$m$ is a transversal.

$\therefore \angle 1 = \angle 3 =30^o$ ....

$(2)$ [Corresponding angles]

$\angle a = \angle 3$ [Vertically opposite angles]

$\therefore \angle a =30^o$ [using

$2$ ]
In addition,

$p$ parallel to

$q$ and

$l$ is a transversal.

$\therefore \angle 2 =30^o$ ... [Corresponding angles]
Here,

$l$ is a straight line.

$\therefore \angle c + \angle 2=180^o$ [Linear pair]

$\therefore 30^o + \angle 2=180^o$ [by equation

$(2)$ ]

$\Rightarrow \angle c=180^o-30^o$

$\Rightarrow \angle c=150^o$
Therefore,

$\Rightarrow \angle a=30^o$

$\Rightarrow \angle b=150^o$

$\Rightarrow \angle c=150^o$

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

Linear pairs, adjacent angles, supplementary angles, and vertically opposite angles, total of

$4$ types of angles will be formed

Prove that the bisectors of a pair of vertically opposite angles are on the same straight line.

Let

$AB$ and

$CD$ be straight lines intersecting at

$O$ .

Also, let

$OX$ be the bisector of

$\angle AOC$ and

$OY$ be the bisector of

$\angle BOD$

$OY$ is the bisector of

$\angle BOD$

$\therefore \ \angle 1 = \angle 6$ ....(i)

$OX$ is the bisector of

$\angle AOC$

$\therefore \ \angle 3 = \angle 4$ .....(ii)

$\angle 2$ and

$\angle 5$ are vertically opposite angles.

$\therefore \ \angle 2 = \angle 5$ .....(iii)

We know that, sum of all angles

$=360^o$

$\therefore\ \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=360^o$

Using the relations from (i), (ii) and (iii), we get:

$\angle 1+\angle 2+\angle 3+\angle 3+\angle 2+\angle 1=360^o$

$2\angle 1+2\angle 2+ 2\angle 3=360^o$

$\Rightarrow \angle DOY+\angle AOD+\angle AOX=180^o$

But,

$\angle DOY+\angle AOD+\angle AOX=\angle XOY$

$\therefore \ \angle XOY=180^o$

Since,

$\angle XOY=180^o$ , both

$OX$ and

$OY$ are on the same straight line.

$\quad \quad [Hence\ proved]$

Name all the pairs of vertically opposite angles in the following figure.

For intersecting lines

$AC$ and

$BD$ ,

$\angle AOB$ and

$\angle DOC$ are vertically opposite angles.

For intersecting lines

$AC$ and

$PQ$ ,

$\angle AOP$ and

$\angle QOC$ are vertically opposite angles.

For intersecting lines

$PQ$ and

$BD$ ,

$\angle POB$ and

$\angle DOQ$ are vertically opposite angles.

In the figure, if lines $PQ$ and $RS$ intersect at point $T$ , such that $\angle PRT={ 40 }^{\circ }, \angle RPT={ 95 }^{\circ }$ and $\angle TSQ={ 75 }^{\circ }$ , find $\angle SQT$ .

Given,

$\angle PRT = 40^{\circ}, \angle RPT = 95^{\circ} , \angle TSQ = 75^{\circ}$
According to the question,

$\angle PRT +\angle RPT + \angle PTR = 180^{\circ}$ (Sum of the interior angles of the triangle.)

$\Rightarrow 40^{\circ} + 95^{\circ} + \angle PTR = 180^{\circ}$

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

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

$\angle PTR = \angle STQ = 45^{\circ}$ (Vertically opposite angles.)
Now,

$\angle TSQ+\angle PTR+\angle SQT=180^{\circ}$ (Sum of the interior angles of the triangle.)

$75^{\circ}+ 45^{\circ} + \angle SQT = 180^{\circ}$

$\Rightarrow 120^{\circ} + \angle SQT = 180^{\circ}$

$\Rightarrow \angle SQT = 60^{\circ}$

Show that if a straight line is perpendicular to one of the two or more parallel lines, then it is also perpendicular to the remaining lines.

Let us assume that

$AB\parallel CD$

$PQ$ is perpendicular to

$AB$

We have to prove that

$PQ$ is perpendicular to

$CD$

$\angle PQB=90^o$ (given)

$\therefore \angle QRD=\angle PQB=90^o$

Corresponding angles of parallel lines

$\therefore PQ$ is perpendicular to

$CD$

In the Fig. lines PQ and RS intersect each other at point O. If $\angle POR : \angle ROQ = 5 : 7$ , Find all the angles.

$\angle POR + \angle ROQ = 180^o$ (Linear pair of angles)

But

$\angle POR : \angle ROQ = 5 : 7$ (Given)

Therefore,

$\angle POR = \dfrac{5}{12} \times 180 = 75^o$

Similarly,

$\angle ROQ = \dfrac{7}{12} \times 180 = 105^o$

Now,

$\angle POS = \angle ROQ = 105^o$ (Vertically opposite angles)

and

$\angle SOQ = \angle POR = 75^o$ (Vertically opposite angles)

$P$ is a point on the bisector of $\angle ABC$ . The line through $P$ parallel to $BA$ meets $BC$ at $Q$ . Prove that $\Delta BPQ$ is an isosceles triangle.

$\angle ABP=\angle PBC$

From the figure,

$\therefore \angle 3=\angle 2$

$\implies PQ||BC$

$\therefore \angle 1=\angle 2$ ....... (Alternate interior angles)

$\angle 1=\angle 3 \{ \angle 2=\angle 3\}$

In on isosceles triangle and the base are same.

$\angle 1=\angle 3$

$\therefore \triangle BPQ$ is an isosceles triangle.

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In the given figure, two lines $AB$ and $CD$ intersect at $O$ . Ray $OE$ is perpendicular to the line $AB$ . If $a:b=3:7,$ then find reflex $\angle BOD$ .

In the given figure,

We have,

$a:b=3:7$

$\angle EOA=90^o$

Let

$a=3x , b=7x$

Since,

$\angle EOC+\angle BOC +\angle EOA=180^o$

$3x+7x+90^o=180^o$

$x=9^o$

Thus

$a=27^o$ and

$b=63^o$

Now,

$=\angle AOB+\angle BOD$

$=63^o+\angle BOD=180^o$

$=\angle BOD=117^o$

In the given figure if line segment $PQ$ and $RS$ intersect at point $T$ , such that $\angle PRT\, = \,40^\circ \,,\,\,\angle RPT\, = \,95^\circ \,\,and\,\,\angle TSQ\, = \,75^\circ \,,\,find\,\,\angle SQT$ .

$In\,RPT$

$\angle TRP + \angle RPT + \angle PTR = 180$

$\angle RTD = 180 - \left( {95 + 40} \right)$

$180 - \left( {135} \right)$

$45$

$\angle RTP = \angle STQ\,\left( {vertically\,opposite\,angle} \right)$

$STQ = 45$

$Now\,in\,\,STQ$

$\angle STQ + \angle TSQ + \angle SQT = 180$

$\angle SQT = 180 - \left( {45 + 75} \right)$

$180 - 120$

$\angle SQT = 60$

Two lines intersecting each other, prove that vertically opposite angles are equal.

If two lines i.e. AB & CD intersect each other. They have two pair of opp. angles.

i.e.

$\angle AOC , \angle DOB, \angle AOB, \angle COB.$

To prove :-

$\angle AOC = \angle DOB$ &

$\angle AOD = \angle COB$

Proof :-

$\angle AOC + \angle AOD = 180^{\circ}$ [linear pair] ___(1)

$\angle AOD + \angle BOD = 180^{\circ}$ [linear pair] ___(2)

from eq. (1) & (2)

$\angle AOC + \angle AOD = \angle AOD + \angle BOD$

i.e.

$[\angle AOC = \angle BOD]$

Similarly

$[\angle AOD = \angle COB]$ Hence proved.

1456810_1040338_ans_c7a2594f64d0400ba80a6c5b6b0e4322.png

In the given figure two lines $AB$ and $CD$ intersect at $O,$ $OE$ is any ray such that $\angle BOE=80^{\circ}$ . If $\angle BOD=30^{\circ}$ , then the find $\angle EOC$ .

$\angle BOE+\angle AOE=180^{\circ}$ (Supplementary Angles )

$\angle AOE=180^{\circ}-80^{\circ}=100^{\circ}$

and

$\angle BOD=\angle AOC=30^{\circ}$ (Vertically Opposite Angles)

Hence,

$\angle EOC=100^{\circ}+30^{\circ}=130^{\circ}$

Write $4$ pairs of vertically opposite angles from the given figure

As shown in fig.

$\angle 1=\angle 4$ and $\angle 2=\angle 3$ (vertically opposite angles)

$\angle 7=\angle 8$ and $\angle 5=\angle 6$ (vertically opposite angles)

Find the measure of $b$ in the following figures.

$\textbf{Step -1: Given figure}$

$\text{Given figure is,}$

$\text{Here }43^{\circ}\text{ and }b\text{ are the vertically opposite angles of the given intersecting straight lines.}$

$\textbf{Step -2: Apply property of angles}$

$\text{So, they are equal}\textbf{ (because vertically opposite angles are equal )}$

$\therefore{b}=43^{\circ}$

$\textbf{Hence the value of b is }\mathbf{43^{\circ}.}$

In figure the value of $x$ , hence find all three indicated angles.

From figure,

$5x=90^o$

$\implies x=18$

Therefore indicated angles,

$3x=3\times18=54^o$
$2x=2\times18=36^0$
$5x=5\times18=90^o$

Find $\dfrac{x+u+y+z}{50}$

$x=z\\u=y..............(1)$

(Vertically opposite Angles)

Now, EF is a straight line

So,

$35^o+y+20^o+z=180^o\Rightarrow y+z=180^o-55^o\Rightarrow y+z=125^o$

Hence,

$\dfrac{x+u+y+z}{50}=\dfrac{z+y+y+z}{50}=\dfrac{2(y+z)}{50}=\dfrac{2*125^o}{50}=5^o$

find the value of x in each of the following figures.

R.E.F image

Solution :-

we have

$83^{\circ}+(6x+19) = 180$ [sum of liner pair]

$6x+102 = 180$

$\Rightarrow 6x = 78$

$\Rightarrow x = 13$

By vertically opposite Angle

$3x+2 = 62$

$\Rightarrow 3x = 60$

$\Rightarrow x = 20$

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Observe the following figure and write each:
(i) Linear pair
(ii) Pair of vertically opposite angles

Linear pairs are,

$\angle 1$ and

$\angle 2$

$\angle 1$ and

$\angle 4$

$\angle 3$ and

$\angle 2$

$\angle 3$ and

$\angle 4$

$\angle 5$ and

$\angle 6$

$\angle 5$ and

$\angle 7$

$\angle 8$ and

$\angle 7$

$\angle 8$ and

$\angle 6$

Pair of vertically opposite angles are,

$\angle 1$ and

$\angle 3$

$\angle 2$ and

$\angle 4$

$\angle 6$ and

$\angle 7$

$\angle 5$ and

$\angle 8$

Fill in the blanks:
_____pair of vertically opposite angle are formed by the intersection of $2$ lines

1901685_1302508_ans_a6bd090adc32479db25ef1479ac1729d.jpg

In the adjoining figure, identify the vertically opposite angles.

The pairs of vertically opposite angles are:

$\angle 1, \angle 3$

and

$\angle 2, \angle 4$

and

$\angle 6, \angle 8$

and

$\angle 5, \angle 7$

In the given figure, lines $AB$ and $CD$ intersect at $O$ .If $\angle{AOC}+\angle{BOE}={70}^{\circ}$ and $\angle{BOD}={40}^{\circ}$ find $\angle{BOE}$ and reflex of $\angle{COE}$

Given:

$\angle{BOD}={40}^{\circ}$

Since

$AB$ and

$CD$ intersects,

$\angle{AOC}=\angle{BOD}$ (vertically opposite angles)

$\angle{AOC}={40}^{\circ}$

Also,

$\angle{AOC}+\angle{BOE}={70}^{\circ}$

$\Rightarrow \angle{BOE}={70}^{\circ}-\angle{AOC}={70}^{\circ}-{40}^{\circ}={30}^{\circ}$

We need to find reflex

$\angle{COE}$

Reflex

$\angle{COE}={360}^{\circ}-\angle{COE}$

Now,

$\angle{AOC}+\angle{COE}+\angle{BOE}={180}^{\circ}$

$\Rightarrow \angle{COE}+\left(\angle{AOC}+\angle{BOE}\right)={180}^{\circ}$

$\Rightarrow \angle{COE}+\left({40}^{\circ}+{30}^{\circ}\right)={180}^{\circ}$

$\Rightarrow \angle{COE}={180}^{\circ}-{70}^{\circ}={110}^{\circ}$

Reflex

$\angle{COE}={360}^{\circ}-{110}^{\circ}={250}^{\circ}$

In Fig., $AB\parallel CD$ and $\angle 1$ and $\angle 2$ are in the ratio 3:Determine all angles from 1 to 8.

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In Fig., PQ and RS intersect each other at point O. If $\angle POR : \angle ROQ = 5:7,$ find all the angles.

We have.

$\angle POR : \angle ROQ = 5 : 7$

So, Let

$\angle POR = 5x$ and

$\angle ROQ = 7x$

Since

$\angle POR$ and

$\angle ROQ$ form a linear pair.

$\therefore \angle POR + \angle ROQ = 180^{\circ}\Rightarrow 5x+7x = 180^{\circ} \Rightarrow x = 15^{\circ}$

$\therefore \angle POR = \left ( 5 \times 15 \right )^{\circ} = 75^{\circ}$ and

$\angle ROQ = \left ( 7 \times 15 \right )^{\circ} = 105^{\circ}$

But,

$\angle QOS = \angle POR$ and

$\angle POS = \angle ROQ$ [vertically opposite angles]

$\therefore \angle QOS = 75^{\circ}$ and

$\angle POS = 105^{\circ}$

In figure, write another name for $\angle 3$ .

$\angle 3$ is nothing but

$\angle AOC.$

In figure, OE is the bisector of $\angle BOD$ . If $\angle 1=70^o$ , find the magnitudes of $\angle 2, \angle 3$ and $\angle 4$ .

Let

$\angle EOB =x$

Given,

$\angle EOB = \angle EOD$ [ ∵ OE is bisector of

$\angle BOD$ ]

$x = \angle 1$ .

$\therefore \angle 1 = x = 70^o$ .

Also,

$\angle BOD = \angle 1 + x = (70+70)^o$

$\angle BOD = 140^o$ .

According to the figure,

$\angle COB = \angle AOD$ and

$\angle AOC = \angle BOD = 140^o$ [ ∵ vertically opposite angles]

We know that,

sum of all the angles around a point

$= 360^o$

$\angle BOD + \angle AOD +\angle AOC + \angle COB = 360^o$

$\angle BOD + \angle AOD +\angle BOD + \angle AOD = 360^o$

$2 \angle BOD + 2 \angle AOD = 360^ o$

$\angle BOD + \angle AOD = 180^ o$

$140^o +\angle AOD = 180^ o$

$\angle AOD = 180^o-140^ o$

$\angle AOD = 40^ o$

$= \angle COB$

$\angle AOC = \angle BOD = 140^o$

$\therefore \angle 2 = \angle 4 = 40^o$ and

$\angle 3 = 140^o$ .

Find the value of x in the following figure.

According to the figure,

$3x^o = 105^o$ [ ∵ vertically opposite angles are equal]

$x^o = \dfrac{105}{3}$

$\therefore x = 35^o$

In figure, lines $l_1$ and $l_2$ intersect at $O$ , forming angles as shown in the figure. If $x=45^o$ , find the values of $y, z$ and $u$ .

Given,

$x=45^o$

As per the diagram, two lines

$l_1$ &

$l_2$ intersected to form two pairs of vertically opposite angles.

$x = z$ -- (1)

$u = y$ --(2)

From (1)

$x = z=45^o$ .

We also know that the sum of all the angles around a point is

$360^o$

$x+u+z+y=360^o$

$\Rightarrow 45+u+45+u=360^o$ {

$\because y=u$ as they are vertically oppposite angles}

$\Rightarrow 90+2u=360^o$

$\Rightarrow 2u=270^o$

$\Rightarrow u=135^o$

$\Rightarrow y=135^o$

$\therefore$

$x = 45^o, z=45^o, u=135^o$ and

$y=135^o$ .

In the given figure, the corresponding arms of $\angle ABC$ and $\angle DEF$ are parallel. If $\angle ABC=75^o$ , then find $\angle DEF$ .

Given: Corresponding arms of

$\angle ABC$ &

$\angle DEF$ are parallel.

Let

$M$ be the point of intersection of angles.

$\therefore \, line \, AB \, || \, line \, DE$ &

$line \, BC \, || \, line \, EF$ .

$\therefore \angle ABC$ &

$\angle DMC$ are corresponding angles.

And, corresponding angles on parallel lines are congruent.

$\therefore \angle ABC = \angle DMC$

$\therefore \angle DMC = 75^{\circ}$ ___(1)

Also,

$\angle DMC$ &

$\angle DEF$ are corresponding angles. {

$MC\parallel EF$ }

$\therefore \angle DMC = \angle DEF$

$\therefore \angle DEF = 75^{\circ}$ ____from (1)

2064431_1677603_ans_e9d0a5e5095e4caab8e5d2eec087e698.png

In figure, three coplanar lines intersect at a point $O$ , forming angles as shown in the figure. Find the values of $x, y, z$ and $u$ .

According to the figure,

$\angle EOB, \angle BOD$ and

$\angle DOF$ are the angles on a straight line.

thus

$\angle EOB + \angle BOD + \angle DOF =180^o$

$x + 90^o +50^o =180^o$

$x = 40^o$

Vertically opposite angles are equal

$z=90^o$

$y=50^o$

$x=u$

$\therefore z=90^o, y=50^o$ and

$x = u = 40^o$

In figure, find the values of $x, y$ and $z$ .

As per the diagram there are two pairs of vertically opposite angles.

and we know that vertically opposite angles are equal.

$y = 25^o$

$x^o = z^o$

and we know that. sum of all the angles around a point is

$360^o$ .

$y^o+x^o+25^o+z^o=360^o$

$25^o+x^o+25^o+x^o=360^o$

$50^o+2x^o=360^o$

$25^o+x^o=180^o$

$x^o=155^o$

$\therefore$

$x=155^o$ ,

$y=25^o$ and

$z=155^o$ ,

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

Let us consider

$AB$ and

$CD$ as the two lines intersecting at a point

$O$ where

$OE$ is the ray bisecting

$\angle BOD$ and

$OF$ is the ray bisecting

$\angle AOC$

To Prove:

$\angle AOF=\angle COF$

Proof:

$EF$ is a straight line passing through the point

$O$ where

$\overrightarrow { OE }$ and

$\overrightarrow { OF }$ are tow opposite rays

From the figure we know that

$\angle AOF$ and

$\angle BOE$ ,

$\angle COF$ and

$\angle DOE$ are vertically opposite angles

so it can be written as

$\angle AOF=\angle BOE$ and

$\angle COF=\angle DOE$

it is given that

$\angle BOE=\angle DOE$

so we can write it as

$\angle AOF=\angle COF$

therefore it is proved that

$\angle AOF=\angle COF$

Two lines $AB$ and $CD$ intersect at $O$ . If $\angle AOC={50}^{o}$ , find $\angle AOD, \angle BOD$ and $\angle BOC$

From the figure we know that

$\angle AOC$ and

$\angle AOD$ from a linear pair

It can also be written as

$\angle AOC+\angle AOD={ 180 }^{ o }$

By substituting the values

${50}^{o}+\angle AOD={ 180 }^{ o }$

$\angle AOD={ 180 }^{ o }-{50}^{o}$

$\angle AOD={130}^{o}$

According to the figure we know that

$\angle AOD$ and

$\angle BOC$ are vertically opposite angles

So we get

$\angle AOD=\angle BOC={ 180 }^{ o }$

According to the figure we know that

$\angle AOC$ and

$\angle BOD$ are vertically opposite angles

So we get

$\angle AOC+\angle BOD={50}^{o}$

In the figure, three coplanar lines $AB,CD$ and $EF$ intersect at a point $O$ , forming angles as shown. Find the values of $x,y,z$ and $t$

From the figure we know that

$\angle COE$ and

$\angle DOF$ are vertically opposite angles

$\angle COE=\angle DOF=\angle z={50}^{o}$

From the figure we know that

$\angle BOD$ and

$\angle COA$ are vertically opposite angles

$\angle BOD=\angle COA=\angle t={90}^{o}$

We also know that

$\angle COA$ and

$\angle AOD$ form a linear pair

$\angle COA+\angle AOD={ 180 }^{ o }$

It can also be written as

$\angle COA+\angle AOF+\angle FOD={ 180 }^{ o }$

By substituting the values we get

${90}^{o}+{x}^{o}+{50}^{o}={ 180 }^{ o }$

${x}^{o}={40}^{o}$

From the figure we know that

$\angle EOB$ and

$\angle AOF$ are vertically opposite angles

$\angle EOB=\angle AOF=x=y=40$

Therefore the values of

$x,y,z,t$ are

$40^{o},40^{o},50^{o},90^{o}$

Two lines $AB$ and $CD$ intersect at a point $O$ such that $\angle BOC+\angle AOD={280}^{o}$ , as shown in the figure. Find all four angles.

From the figure we know that

$\angle AOD$ and

$\angle BOC$ are vertically opposite angles

$\angle AOD=\angle BOC$

It is given that

$\angle BOC+\angle AOC={ 280 }^{ o }$

we know that

$\angle AOD=\angle BOC$

so it can be written as

$\angle AOD+\angle AOD={ 180 }^{ o }$

$2\angle AOD={ 280 }^{ o }$

$\angle AOD=\dfrac{180}{2}$

$\angle AOD=\angle BOC={ 140 }^{ o }$

From the figure we know that

$\angle AOC$ and

$\angle AOC$ form a linear pair

So it can be written as

$\angle AOC+\angle AOD={ 180 }^{ o }$

substituting the values

$\angle AOC+{ 140 }^{ o }={ 180 }^{ o }$

$\angle AOC={40}^{o}$

From the figure we know that

$\angle AOC$ and

$\angle BOD$ are vertically opposite angles

$\angle AOC=\angle BOD={40}^{o}$

Therefore

$\angle AOC={40}^{o}$ ,

$\angle BOC={ 140 }^{ o }$ ,

$\angle AOD={ 140 }^{ o }$ ,

$\angle BOD={ 40 }^{ o }$

If two straight lines intersect each other in such a way that one of the angles formed measures ${90}^{o}$ , show that each of the remaining angles measures ${90}^{o}$ .

Consider two lines

$AB$ and

$CD$ intersecting at a point

$O$ with

$\angle AOC={90}^{o}$

From the figure we know that

$\angle AOC$ and

$\angle BOD$ are vertically opposite angles

so it can be written as

$\angle AOC+\angle AOD={ 180 }^{ o }$

substituting the values

${90}^{o}+\angle AOD={ 180 }^{ o }$

$\angle AOD={ 180 }^{ o }-{ 90 }^{ o }$

$\angle AOD={ 90 }^{ o }$

from the figure we also know that

$\angle BOC=\angle AOD$ are vertically opposite angles

$\angle BOC=\angle AOD={90}^{o}$

Therefore, it is proved that each of the remaining angle is

${90}^{o}$

1564469_1715109_ans_269ce85bd1764ae9a3d889ab8bb08856.png

In the given figure, the two lines $AB$ and $CD$ intersect at a point $O$ such that $\angle BOC={125}^{o}$ . Find the values of $x,y$ and $z$

From the figure we know that

$\angle AOC$ and

$\angle BOC$ form a linear pair of angles

so it can be written as

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

by substituting the values we get

$x+{125}^{o}={ 180 }^{ o }$

$x={55}^{o}$

from the figure we know that

$\angle AOD$ and

$\angle BOC$ are vertically opposite angles

so we get

$y={125}^{o}$

from the figure we know that

$\angle BOD$ and

$\angle AOC$ are vertically opposite angles

so we get

$z={55}^{o}$

therefore the values of

$x,y,z$ are

${55}^{o},{125}^{o},{55}^{o}$ respectively.

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the pairs of vertically opposite angles.

Given

$AB$ and

$CD$ intersect each other at point

$C$

$\angle AOC=68^o$
Here

$AOB$ is a line
The pairs of vertically opposite angles are.

$\angle AOC$ and

$\angle BOD$

$\angle BOC$ and

$\angle DOA$

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the pairs of adjacent angles.

Given

$AB$ and

$CD$ intersect each other at point

$C$

$\angle AOC=68^o$
Here

$AOB$ is a line
The pairs of adjacent angles are.

$\angle AOC$ and

$\angle BOC,$

$\angle BOC$ and

$\angle BOD,$

$\angle BOD$ and

$\angle AOD$

$\angle AOD$ and

$\angle AOC$

In the figure, three lines $AB,CD$ and $EF$ intersect at a point $O$ such that $\angle AOE={35}^{o}$ and $\angle BOD={40}^{o}$ . Find the measure of $\angle AOC,\angle BOF,\angle COF$ and $\angle DOE$

It is given that

$\angle BOD={ 180 }^{ o }$

from the figure we know that

$\angle BOD$ and

$\angle AOC$ are vertically opposite angles

$\angle AOC=\angle BOD={40}^{o}$

It is given that

$\angle AOE={35}^{o}$

from the figure we know that

$\angle BOF$ and

$\angle AOE$ are vertically opposite angles

$\angle AOE=\angle BOF={35}^{o}$

from the figure, we know that

$AOB$ is a straight line

so it can be written as

$\angle AOB={ 180 }^{ o }$

we can write it as

$\angle AOE+\angle EOD+\angle BOD={ 180 }^{ o }$

by substituting the values

${35}^{o}+\angle EOD+{40}^{o}={ 180 }^{ o }$

$\angle EOD={105}^{o}$

from the figure, we know that

$\angle COF$ and

$\angle EOD$ are

vertically opposite angles

$\angle COF=\angle EOD={105}^{o}$

In the adjoining figure, it is being given that $AO \parallel CD, OB \parallel CE$ and $\angle AOC=50^o$ . Find the measure of $\angle ECD$ .

As,

$AO\parallel CD,$

$\angle AOD= \angle CDB = 50^o$ [Corresponding angles]

Also,

$OB\parallel CE$ and sum of interior angles on the same side of transversal are supplementary so,

$\angle ECD+ \angle CDB=180^o$

$\Rightarrow \angle ECD + 50^o = 180^o$

$\Rightarrow \angle ECD=180^o-50^o$

$\Rightarrow \angle ECD=130^o$

The legs of a stool make an angle of $35^o$ with the floor as shown in
Fig. Find the angle $x$ and $y$

Given,

$l$ parallel to

$m$ and

$QR$ is a transversal.

$\therefore \angle x=35^o$ [Alternative interior angles]

$\therefore \angle y+35^o=180^o$ [Co-interior angles]

$\Rightarrow \angle y=180^o-35^o$

$\Rightarrow \angle y=145^o$
Therefore, the angle are

$\therefore \angle x=35^o$

$\angle y=145^o$

1669975_1791101_ans_265a9df7bbd1430fbd0bfe71b1f72353.png

Three lines $AB,\ CD$ and $EF$ intersect each other at $O$ . If $\angle AOE=30^o$ and $\angle DOB=40^o$ Fig. find $\angle COF$ .

$AB$ is a straight line.

$\therefore \angle AOE + \angle EOD +\angle DOB=180^o$

$\Rightarrow \angle EOD=180^o-30^o-40^o$

$\Rightarrow \angle EOD=110^o \ .....(1)$

Now,

$CD$ and

$EF$ intersect each other at point

$O$ .

$\therefore \angle COF =\angle EOD =110^o$ [vertically opposite angles]

Therefore,

$\angle COF=110^o$

In Fig., $AB \parallel CD$ . Find the reflex $\angle EFG$ .

In given figure,

$AB$ parallel to

$l$ and

$EF$ is a transversal.

$\therefore \angle 1 =34^o$ ... Alternate interior angles

Now,

$CD$ parallel to

$l$ and

$FG$ is a transversal.

$\therefore \angle 2+135^o=180^o$ ...Co-interior angles

$\Rightarrow \angle 2 =180^o-135^o$

$\Rightarrow \angle 2 =45^o$

Also,

$\angle EFG = \angle 1 + \angle 2 =34^o+45^o$

$\Rightarrow \angle EFG=79^o$

The reflex of

$\angle EFG$

$\Rightarrow \angle EFG=360^o-79^o$

$\Rightarrow \angle EFG=281^o$

In Fig. , If $RP =RQ$ , find the value of $x$

Given,

$RQ=RP$

$\Rightarrow \angle ROP=\angle RPQ=x$ [

$\because$ Angle opposite to equal side is equal]

Now,

$\angle RPQ=50^{o}$ [Vertically opposite angles]

$\therefore x=50^{o}$

Angles formed by two intersecting lines having no common arms are called ..

Angles formed by two intersecting lines having no common arms are called vertically opposite angles

If two lines intersect and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are ..

If two lines intersect and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles.

Find the value of x in each of the following figures.

$\Delta$ ABC

$CA=CB$

$\angle A=\angle{ABC}=x$

$\angle{EBD}=36^\circ$

$\angle{ABC}=\angle{EBD}=36^\circ$ (Vertically opposite angles)

$x=36^\circ$
1774327_1803202_ans_ab378c4f4d644c5ebae739b409759489.jpg

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Find the value of $\angle a$ .

Given

$AB$ and

$CD$ intersect each other at point

$C$

$\angle AOC=68^o$
Here

$AOB$ is a line
Hence,

$\angle AOC+\angle BOC =180^o$ [Linear pair of angles]

$68^o +a=180^o$

$a=180^o-68^o$

$a=112^o$

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find the value of $x$ .

By referring given figure,

The value of

$x$ is calculated as,

$3x + 15^\circ = 4x - 5^\circ$ (By vertically opposite angles)

$3x - 4x = -5^\circ - 15^\circ$

$-x = -20^\circ$

$x = 20^\circ$

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find $\angle BPD$ .

By referring given figure,

The value of

$x$ is calculated as,

$3x + 15^\circ = 4x - 5^\circ$ (By vertically opposite angles)

$3x - 4x = -5^\circ - 15^\circ$

$-x = -20^\circ$

$x = 20^\circ$

The value of

$\angle BPD$ is calculated as,

The sum of adjacent angles is

$180^∘$

$\angle BPD+ \angle BPC = 180^\circ$

$\angle BPD = 180^\circ - \angle BPC$

$= 180^\circ - (4x - 5^\circ)$

$= 180^\circ - (4 \times 20^\circ - 5^\circ )$

$= 180^\circ - 80^\circ + 5^\circ$

$= 105^\circ$

Two straight lines AB and CD intersect each other at a point O and angle $AOC = 50^\circ,$ find:
angle BOD.

Given

$\angle AOC = 50^\circ$

We know that,

Vertically opposite angles are equal

So,

$\angle BOD = \angle AOC$

Therefore,

$\angle BOD = 50^\circ$

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ then find $\angle BPC$ .

By referring the given figure, the value of

$x$ is calculated as,

$3x + 15^\circ = 4x - 5^\circ$ ( Vertically opposite angles )

$3x - 4x = -5^\circ - 15^\circ$

$-x = -20^\circ$

$x = 20^\circ$

The value of

$\angle BPC$ is calculated as,

$\angle BPC = (4x - 5^\circ)$

$= (4 \times 20^\circ - 5^\circ)$

$= 80^\circ - 5^\circ$

$= 75^\circ$

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 $\angle 4$

$\angle 2$ and

$\angle 4$ = Adjacent angles

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the reflex angles formed and write the measures of each.

Given

$AB$ and

$CD$ intersect each other at point

$C$

$\angle AOC=68^o$
Here

$AOB$ is a line
The reflex angles in the given figure are
reflex

$\angle AOD=$ reflex

$\angle BOC$
reflex

$\angle BOD=$ reflex

$\angle AOC$

reflex

$\angle AOD=\angle AOC+\angle COD=68^\circ+180^\circ=248^\circ$
similarly, reflex

$\angle BOC=248^\circ$

$\angle AOC+\angle BOC=180^\circ$ [linear pair]

$\Rightarrow 68^\circ+a=180^\circ$

$\Rightarrow a=180^\circ-68^\circ\\\therefore\,a=112^\circ$

$\angle AOD=\angle BOC$ [vertically opposite angle]

$\angle AOD=a=112^\circ$

reflex

$\angle BOD=\angle BOC+\angle AOC+\angle AOD=a+68^\circ+a=112^\circ+68^\circ+112^\circ=292^\circ$
similarly, reflex

$AOC=292^\circ$

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find $\angle APD$ .

By referring given figure,

The value of

$x$ is calculated as,

$3x + 15^\circ = 4x - 5^\circ$ (By vertically opposite angles)

$3x - 4x = -5^\circ - 15^\circ$

$-x = -20^\circ$

$x = 20^\circ$

the value of

$\angle APD$ is calculated as,

$\angle APD = 3x + 15^\circ$

$= 3 \times 20^\circ + 15^\circ$

$= 60^\circ + 15^\circ$

$= 75^\circ$

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.
$\angle 1$ and $\angle 10$

$\angle 1$ and

$\angle 10$ = Corresponding angles

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.
$\angle 4$ and $\angle 5$

$\angle 4$ and

$\angle 5$ = Alternate interior angles

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.
$\angle 6$ and $\angle 12$

$\angle 6$ and

$\angle 12$ = Alternate exterior angles

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:

$\angle 4 \quad and \quad \angle 6$

$\angle 4 \quad and \quad \angle 6$ are allied or co -interior angles.

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:

$\angle 7 \quad and \quad \angle 13$

$\angle 7 \quad and \quad \angle 13$ are corresponding angles.

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:

$\angle 1 \quad and \quad \angle 8$

$\angle 1 \quad and \quad \angle 8$ are exterior alternate angles.

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:

$\angle 5 \quad and \quad \angle 7$

$\angle 5 \quad and \quad \angle 7$ are adjacent angles.

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:

$\angle 6 \quad and \quad \angle 8$

$\angle 6 \quad and \quad \angle 8$ are opposite angles.

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:

$\angle 10 \quad and \quad \angle 12$

$\angle 10 \quad and \quad \angle 12$ are opposite angles.

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:

$\angle 3 \quad and \quad \angle 6$

$\angle 3 \quad and \quad \angle 6$ are interior alternate angles.

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:

$\angle 1 \quad and \quad \angle 5$

$\angle 1 \quad and \quad \angle 5$ are corresponding angles.

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:

$\angle 4 \quad and \quad \angle 6$

$\angle 4 \quad and \quad \angle 6$ are allied or co-interior angles.

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:

$\angle 2 \quad and \quad \angle 13$

$\angle 2 \quad and \quad \angle 13$ are allied or co-interior angles.

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:

$\angle 4 \quad and \quad \angle 5$

$\angle 4 \quad and \quad \angle 5$ are adjacent angles.

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:

$\angle 7 \quad and \quad \angle 9$

$\angle 7 \quad and \quad \angle 9$ are opposite angles.

In the given figures, the directed lines are parallel to each other. Find the unknown angles.

If the lines are parallel

$a = b$ are corresponding angles

$a =c$ are vertically opposite angles

$a = b =c$
Here

$b = 60^0$ are vertically opposite angles
Therefore

$a =b= c = 60^0$

In the above figure, lines $AB$ and $CD$ intersect at $O$ . If $\angle AOC + \angle BOE = 70^{\circ} and \angle BOD = 40^{\circ}$ , find $\angle BOE$ and reflex $\angle COE$

$\angle AOC = \angle BOD = 40^{\circ}$ .......Vertically opposite angles

$\angle BOD = 40^{\circ}$ (Date)

$\angle AOC = \angle BOD = 40^{\circ}$ ........... Vertically opposite angles

$\therefore \angle AOC + \angle BOE = 70^{\circ}$

$40^{\circ} + \angle BOE = 70^{\circ}$

$\therefore \angle BOE = 70^{\circ} - 40^{\circ} \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ = 30^{\circ}$

$\angle AOD = 180^{\circ} - \angle BOD \\= 180^{\circ} - 40^{\circ} \\= 140^{\circ}$
Reflex angle

$COE = \angle COA + \angle AOD + \angle BOD + \angle BOE$

$= 40^{\circ} + 140^{\circ} + 40^{\circ} + 30^{\circ}$

$\therefore$ Reflex angle ,

$\angle COE = 250^{\circ}$
1829918_1869591_ans_6689cc27277b4870b9927a5a9202f36d.PNG

Two lines are perpendicular to two parallel lines respectively. Show that these two lines are also parallel.

$\because$

$AB$ and

$CD$ are perpendicular to

$I$ and

$m$

$\therefore \angle APQ = 90^o$ and

$\angle CQM = 90^o$

$\angle APQ = \angle CQM = 90^o$ (Corresponding - angles)

$\therefore$ Hence

$AB || CD$

Find the magnitude of $\angle BOD$

$\angle AOC = 52^o$ (given)

$\angle AOC=\angle BOD$ (Vertically opposite angles)

$\therefore \,\,\,\ \angle BOD = 52^o$

In figure, BA || ED and BC || EF. Show that $\angle ABC + \angle DEF = 180^o$

$\because BA || ED$

$\angle 1 = \angle 2$ .......(i)
(Corresponding angles)
Also BC || EF

$\Rightarrow \angle 2 + \angle 3 = 180^o$ .......(ii)
(Sum of the interior angles on the same side of a transversal is

$180^o$ )
From (i) and (ii), we get

$\angle 1 + \angle 3 = 180^o$
i.e.

$\angle ABC + \angle DEF = 180^o$
1867932_1879514_ans_97476a86f8f44c9bae7a9248af61b790.png

In the adjoining figure if lines $\mathrm{PQ}$ and $\mathrm{RS}$ intersect at point $T,$ such that $\angle P R T=40^{\circ},$
$\angle \mathrm{RPT}=95^{\circ}$ and $\angle \mathrm{TSQ}=75^{\circ},$ find $\angle \mathrm{SQT}$

$\ln \Delta \mathrm{PRT}$

$\angle \mathrm{RPT}+\angle \mathrm{PRT}+\angle \mathrm{RTP}=180^{\circ} \quad$ (angle sum property of a triangle)

$\Rightarrow 95^{\circ}+40^{\circ}+\angle \mathrm{RTP}=180^{\circ}$

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

$\Rightarrow \angle R T P=180^{\circ}-135^{\circ}$

$=45^{\circ}$
Now

$\angle \mathrm{RTP}=\angle \mathrm{STQ}=45^{\circ} \quad$ (vertically opposite angles)

$\ln \Delta \mathrm{S} \mathrm{T} \mathrm{Q}$

$\angle S T Q+\angle T S Q+\angle S Q T=180^{\circ} \quad$ (angle sum property of a triangle)

$\Rightarrow 45^{\circ}+75^{\circ}+\angle S Q T=180^{\circ}$

$\Rightarrow \angle S Q T=180^{\circ}-\left(45^{\circ}+75^{\circ}\right)$

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

$\angle S Q T=60^{\circ}$

Find the measure of $x$ in the given figure.

$\textbf{Step -1: Finding the angle sum of the polygon in the figure.}$

$\text{The polygon given in the figure is a 5 sided polygon.}$

$\therefore\text{ Angle Sum}=(5-2)180^{\circ}=540^{\circ}$

$\textbf{Step -2: Finding the value of x.}$

$\text{Since }\angle A\text{ and }100^{\circ}\text{ lie on the same line,}$

$\angle A+100^{\circ}=180^{\circ}$

$\implies \angle A=80^{\circ}$

$\text{Since }\angle B\text{ and }65^{\circ}\text{ lie on the same line,}$

$\angle B+65^{\circ}=180^{\circ}$

$\implies \angle B=115^{\circ}$

$\text{Since }\angle C\text{ and }40^{\circ}\text{ lie on the same line,}$

$\angle C+40^{\circ}=180^{\circ}$

$\implies \angle C=140^{\circ}$

$\text{Since }\angle E\text{ and }45^{\circ}\text{ lie on the same line,}$

$\angle E+45^{\circ}=180^{\circ}$

$\implies \angle E=135^{\circ}$

$\angle A+\angle B+\angle C+\angle D+\angle E=540^{\circ}\textbf{ (Angle Sum of 5 sided polygon)}$

$\implies \angle D+80+115+140+135=540$

$\implies \angle D=70$

$\text{Since }\angle D\text{ and }x\text{ lie on the same line,}$

$\angle D+x=180^{\circ}$

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

$\implies x=110^{\circ}$

$\textbf{Hence, The value of x is }\mathbf{110^{\circ}}.$

In figure, HOPE is a parallelogram , Find the angle measure $x , y$ and $z$ . State the properties used to find them.

2039681_1362041_ans_e9e7c501209e465383fce33459d6c27a.jpg

Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.

Draw two lines

$AB$ and

$CD$ intersecting each other at

$O$ .

We know that the vertically opposite angles are equal.

Therefore,

$∠BOC = ∠AOD$ and

$∠AOC = ∠BOD$ .

We bisect angle

$AOC$ and draw the bisecting ray as

$OX$ .

Similarly, we bisect angle

$BOD$ and draw the bisecting ray as

$OY$ .

Now,

$∠XOA + ∠AOD + ∠DOY$

$= \dfrac {1}{2} ∠AOC + ∠AOD + \dfrac {1}{2} ∠BOD$

$= \dfrac {1}{2} ∠BOD + ∠AOD + \dfrac {1}{2} ∠BOD$

[As,

$∠AOC = ∠BOD$ ]

$= ∠AOD + ∠BOD$

Since,

$AB$ is a line.

Therefore,

$∠AOD$ and

$∠BOD$ are supplementary angles and the sum of these two angles will be

$180^°$ .

Therefore,

$∠XOA + ∠AOD + ∠DOY = 180^°$

We know that the angles on one side of a straight line will always add to

$180^°$ .

Also, the sum of the angles is

$180^°$ .

Therefore,

$XY$ is a straight line.

Thus,

$OX$ and

$OY$ are in the same line.

1424799_1674959_ans_09fa36a9de89431b9632a187daa8a60f.png

In Fig. arms BA and BC of $\angle ABC$ are respectively parallel to arms ED and EF of $\angle DEF$ . Prove that $\angle ABC= \angle DEF$ .

2005724_1669449_ans_a432f130919a4ef38ee2ad2ce126b3ec.jpg

In Fig., line $n$ is a transversal to lines $l$ and $m$ . Identify the following:
Angle alternate to $\angle PQR$ and angle corresponding to angles $\angle RQF$ and angle alternate to $\ \angle PQE$ in Fig.

We know that the corresponding angles are the angles that occupy the same relative position at each intersection where a straight line crosses two others. Also, 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, we have:

Angle alternate to

$∠PQR$ is

$∠QRA$ .

Angle corresponding to

$∠RQF$ is

$∠ARB$ .

Angle alternate to

$∠PQE$ is

$∠ARB$ .

Lines And Angles - Class 7 Maths - Extra Questions

In Fig., lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP+∠CDP=∠DPB\angle ABP+\angle CDP= \angle DPB.

In figure, write another name for ∠2\angle 2.

Find xx and yy.

In the adjoining figure, name the following pairs of angles.Obstuse vertically opposite angles

In the given figure three lines ↔AB,↔CD\overleftrightarrow{AB}, \overleftrightarrow{CD} and ↔EF\overleftrightarrow{EF} intersecting at OO. Find the values of x,yx, y and zz it is being given that x:y:z=2:3:5x : y : z = 2 : 3 : 5

Two complementary angles are in the ratio 4:54 : 5. Find the angles.

In the figure, find the values of xx and yy and then show that AB\parallel CDAB\parallel CD.

Vertically opposite angle of \angle XVW\angle XVW is _______.

Write all pairs of vertically opposite angles.

Vertically opposite angle of \angle XVZ\angle XVZ is __________.

Find x:

How many pair of vertically opposite angles are formed?

Write the pair of vertically opposite angles.

In the given figure , the arms of two angles are parallel. if \angle ABC = {70^ \circ }\angle ABC = {70^ \circ } , then find (i) \angle DGC\angle DGC(ii) \angle DEF\angle DEF

Fill in the blanks:If two line intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are_______.

In the adjoining figure,p\parallel qp\parallel q. Find the unknown angles.

In the given figure, if \angle A O C + \angle B O D = 266 ^ { \circ }\angle A O C + \angle B O D = 266 ^ { \circ } find all the four angles.

In the given figure, find xx.

Corresponding angles are on the _________ side of the transversal.

The sum of two vertically opposite angles is 166^\circ.166^\circ. Find each of the angles.

A road crosses a railway line at an angle of 30^o30^o as shown in Fig.Find the values of a,\ ba,\ b and cc.

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

Prove that the bisectors of a pair of vertically opposite angles are on the same straight line.

Name all the pairs of vertically opposite angles in the following figure.

In the figure, if lines PQPQ and RSRS intersect at point TT, such that \angle PRT={ 40 }^{\circ }, \angle RPT={ 95 }^{\circ }\angle PRT={ 40 }^{\circ }, \angle RPT={ 95 }^{\circ } and \angle TSQ={ 75 }^{\circ }\angle TSQ={ 75 }^{\circ }, find \angle SQT\angle SQT.

Show that if a straight line is perpendicular to one of the two or more parallel lines, then it is also perpendicular to the remaining lines.

In the Fig. lines PQ and RS intersect each other at point O. If \angle POR : \angle ROQ = 5 : 7\angle POR : \angle ROQ = 5 : 7, Find all the angles.

PP is a point on the bisector of \angle ABC\angle ABC. The line through PP parallel to BABA meets BCBC at QQ. Prove that \Delta BPQ\Delta BPQ is an isosceles triangle.

In the given figure, two lines ABAB and CDCD intersect at OO. Ray OEOE is perpendicular to the line ABAB. If a:b=3:7,a:b=3:7, then find reflex \angle BOD\angle BOD.

Two lines intersecting each other, prove that vertically opposite angles are equal.

In the given figure two lines ABAB and CDCD intersect at O,O, OEOE is any ray such that \angle BOE=80^{\circ}\angle BOE=80^{\circ}. If \angle BOD=30^{\circ}\angle BOD=30^{\circ}, then the find \angle EOC\angle EOC.

Write 44 pairs of vertically opposite angles from the given figure

Find the measure of bb in the following figures.

In figure the value of xx, hence find all three indicated angles.

Find \dfrac{x+u+y+z}{50}\dfrac{x+u+y+z}{50}

find the value of x in each of the following figures.

Observe the following figure and write each:(i) Linear pair(ii) Pair of vertically opposite angles

Fill in the blanks:_____pair of vertically opposite angle are formed by the intersection of 22 lines

In the adjoining figure, identify the vertically opposite angles.

In the given figure, lines ABAB and CDCD intersect at OO.If \angle{AOC}+\angle{BOE}={70}^{\circ}\angle{AOC}+\angle{BOE}={70}^{\circ} and \angle{BOD}={40}^{\circ}\angle{BOD}={40}^{\circ} find \angle{BOE}\angle{BOE} and reflex of \angle{COE}\angle{COE}

In Fig., AB\parallel CD AB\parallel CD and \angle 1 \angle 1 and \angle 2 \angle 2 are in the ratio 3:Determine all angles from 1 to 8.

In Fig., PQ and RS intersect each other at point O. If \angle POR : \angle ROQ = 5:7, \angle POR : \angle ROQ = 5:7, find all the angles.

In figure, write another name for \angle 3\angle 3.

In figure, OE is the bisector of \angle BOD\angle BOD. If \angle 1=70^o\angle 1=70^o, find the magnitudes of \angle 2, \angle 3\angle 2, \angle 3 and \angle 4\angle 4.

Find the value of x in the following figure.

In figure, lines l_1l_1 and l_2l_2 intersect at OO, forming angles as shown in the figure. If x=45^ox=45^o, find the values of y, zy, z and uu.

In the given figure, the corresponding arms of \angle ABC\angle ABC and \angle DEF\angle DEF are parallel. If \angle ABC=75^o\angle ABC=75^o, then find \angle DEF\angle DEF.

In figure, three coplanar lines intersect at a point OO, forming angles as shown in the figure. Find the values of x, y, zx, y, z and uu.

In figure, find the values of x, yx, y and zz.

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

Two lines ABAB and CDCD intersect at OO. If \angle AOC={50}^{o}\angle AOC={50}^{o}, find \angle AOD, \angle BOD\angle AOD, \angle BOD and \angle BOC\angle BOC

In the figure, three coplanar lines AB,CDAB,CD and EFEF intersect at a point OO, forming angles as shown. Find the values of x,y,zx,y,z and tt

Two lines ABAB and CDCD intersect at a point OO such that \angle BOC+\angle AOD={280}^{o}\angle BOC+\angle AOD={280}^{o}, as shown in the figure. Find all four angles.

If two straight lines intersect each other in such a way that one of the angles formed measures {90}^{o}{90}^{o}, show that each of the remaining angles measures {90}^{o}{90}^{o}.

In the given figure, the two lines ABAB and CDCD intersect at a point OO such that \angle BOC={125}^{o}\angle BOC={125}^{o}. Find the values of x,yx,y and zz

In the given figure, lines ABAB and CDCD intersect at point OO.Name all the pairs of vertically opposite angles.

In the given figure, lines ABAB and CDCD intersect at point OO.Name all the pairs of adjacent angles.

In the figure, three lines AB,CDAB,CD and EFEF intersect at a point OO such that \angle AOE={35}^{o}\angle AOE={35}^{o} and \angle BOD={40}^{o}\angle BOD={40}^{o}. Find the measure of \angle AOC,\angle BOF,\angle COF\angle AOC,\angle BOF,\angle COF and \angle DOE\angle DOE

In the adjoining figure, it is being given that AO \parallel CD, OB \parallel CE AO \parallel CD, OB \parallel CE and \angle AOC=50^o \angle AOC=50^o . Find the measure of \angle ECD \angle ECD .

The legs of a stool make an angle of 35^o35^o with the floor as shown inFig. Find the angle xx and yy

Three lines AB,\ CDAB,\ CD and EFEF intersect each other at OO. If \angle AOE=30^o\angle AOE=30^o and \angle DOB=40^o\angle DOB=40^o Fig. find \angle COF\angle COF.

In Fig., AB \parallel CDAB \parallel CD. Find the reflex \angle EFG\angle EFG.

In Fig. , If RP =RQRP =RQ, find the value of xx

Angles formed by two intersecting lines having no common arms are called ..

If two lines intersect and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are ..

Find the value of x in each of the following figures.

In the given figure, lines ABAB and CDCD intersect at point OO.Find the value of \angle a\angle a.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If \angle BPC = 4x - 5^\circ\angle BPC = 4x - 5^\circ and \angle APD = 3x + 15^\circ,\angle APD = 3x + 15^\circ, find the value of xx.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If \angle BPC = 4x - 5^\circ\angle BPC = 4x - 5^\circ and \angle APD = 3x + 15^\circ,\angle APD = 3x + 15^\circ, find \angle BPD\angle BPD.

Two straight lines AB and CD intersect each other at a point O and angle AOC = 50^\circ,AOC = 50^\circ, find: angle BOD.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If \angle BPC = 4x - 5^\circ\angle BPC = 4x - 5^\circ and \angle APD = 3x + 15^\circ,\angle APD = 3x + 15^\circ, then find \angle BPC\angle BPC.

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\angle 2 and \angle 4\angle 4

In the given figure, lines ABAB and CDCD intersect at point OO.Name all the reflex angles formed and write the measures of each.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If \angle BPC = 4x - 5^\circ\angle BPC = 4x - 5^\circ and \angle APD = 3x + 15^\circ,\angle APD = 3x + 15^\circ, find \angle APD\angle APD.

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 1\angle 1 and \angle 10\angle 10

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 4\angle 4 and \angle 5\angle 5

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 6\angle 6 and \angle 12\angle 12

In questions 1 1 and 2, 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 6 \angle 4 \quad and \quad \angle 6

In questions 1 1 and 2, 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 7 \quad and \quad \angle 13 \angle 7 \quad and \quad \angle 13

In Fig., lines AB and CD are parallel and P is any point as shown in the figure. Show that $\angle ABP+\angle CDP= \angle DPB$ .

In figure, write another name for $\angle 2$ .

Find $x$ and $y$ .

In the adjoining figure, name the following pairs of angles.
Obstuse vertically opposite angles

In the given figure three lines $\overleftrightarrow{AB}, \overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$ intersecting at $O$ . Find the values of $x, y$ and $z$ it is being given that $x : y : z = 2 : 3 : 5$

Two complementary angles are in the ratio $4 : 5$ . Find the angles.

In the figure, find the values of $x$ and $y$ and then show that $AB\parallel CD$ .

Vertically opposite angle of $\angle XVW$ is _______.

Vertically opposite angle of $\angle XVZ$ is __________.

In the given figure , the arms of two angles are parallel. if $\angle ABC = {70^ \circ }$ , then find
(i) $\angle DGC$
(ii) $\angle DEF$

Fill in the blanks:
If two line intersect at a point and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are_______.

In the adjoining figure, $p\parallel q$ . Find the unknown angles.

In the given figure, if $\angle A O C + \angle B O D = 266 ^ { \circ }$ find all the four angles.

In the given figure, find $x$ .

The sum of two vertically opposite angles is $166^\circ.$ Find each of the angles.

A road crosses a railway line at an angle of $30^o$ as shown in Fig.
Find the values of $a,\ b$ and $c$ .

In the figure, if lines $PQ$ and $RS$ intersect at point $T$ , such that $\angle PRT={ 40 }^{\circ }, \angle RPT={ 95 }^{\circ }$ and $\angle TSQ={ 75 }^{\circ }$ , find $\angle SQT$ .

In the Fig. lines PQ and RS intersect each other at point O. If $\angle POR : \angle ROQ = 5 : 7$ , Find all the angles.

$P$ is a point on the bisector of $\angle ABC$ . The line through $P$ parallel to $BA$ meets $BC$ at $Q$ . Prove that $\Delta BPQ$ is an isosceles triangle.

In the given figure, two lines $AB$ and $CD$ intersect at $O$ . Ray $OE$ is perpendicular to the line $AB$ . If $a:b=3:7,$ then find reflex $\angle BOD$ .

In the given figure two lines $AB$ and $CD$ intersect at $O,$ $OE$ is any ray such that $\angle BOE=80^{\circ}$ . If $\angle BOD=30^{\circ}$ , then the find $\angle EOC$ .

Write $4$ pairs of vertically opposite angles from the given figure

Find the measure of $b$ in the following figures.

In figure the value of $x$ , hence find all three indicated angles.

Find $\dfrac{x+u+y+z}{50}$

Observe the following figure and write each:
(i) Linear pair
(ii) Pair of vertically opposite angles

Fill in the blanks:
_____pair of vertically opposite angle are formed by the intersection of $2$ lines

In the given figure, lines $AB$ and $CD$ intersect at $O$ .If $\angle{AOC}+\angle{BOE}={70}^{\circ}$ and $\angle{BOD}={40}^{\circ}$ find $\angle{BOE}$ and reflex of $\angle{COE}$

In Fig., $AB\parallel CD$ and $\angle 1$ and $\angle 2$ are in the ratio 3:Determine all angles from 1 to 8.

In Fig., PQ and RS intersect each other at point O. If $\angle POR : \angle ROQ = 5:7,$ find all the angles.

In figure, write another name for $\angle 3$ .

In figure, OE is the bisector of $\angle BOD$ . If $\angle 1=70^o$ , find the magnitudes of $\angle 2, \angle 3$ and $\angle 4$ .

In figure, lines $l_1$ and $l_2$ intersect at $O$ , forming angles as shown in the figure. If $x=45^o$ , find the values of $y, z$ and $u$ .

In the given figure, the corresponding arms of $\angle ABC$ and $\angle DEF$ are parallel. If $\angle ABC=75^o$ , then find $\angle DEF$ .

In figure, three coplanar lines intersect at a point $O$ , forming angles as shown in the figure. Find the values of $x, y, z$ and $u$ .

In figure, find the values of $x, y$ and $z$ .

Two lines $AB$ and $CD$ intersect at $O$ . If $\angle AOC={50}^{o}$ , find $\angle AOD, \angle BOD$ and $\angle BOC$

In the figure, three coplanar lines $AB,CD$ and $EF$ intersect at a point $O$ , forming angles as shown. Find the values of $x,y,z$ and $t$

Two lines $AB$ and $CD$ intersect at a point $O$ such that $\angle BOC+\angle AOD={280}^{o}$ , as shown in the figure. Find all four angles.

If two straight lines intersect each other in such a way that one of the angles formed measures ${90}^{o}$ , show that each of the remaining angles measures ${90}^{o}$ .

In the given figure, the two lines $AB$ and $CD$ intersect at a point $O$ such that $\angle BOC={125}^{o}$ . Find the values of $x,y$ and $z$

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the pairs of vertically opposite angles.

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the pairs of adjacent angles.

In the figure, three lines $AB,CD$ and $EF$ intersect at a point $O$ such that $\angle AOE={35}^{o}$ and $\angle BOD={40}^{o}$ . Find the measure of $\angle AOC,\angle BOF,\angle COF$ and $\angle DOE$

In the adjoining figure, it is being given that $AO \parallel CD, OB \parallel CE$ and $\angle AOC=50^o$ . Find the measure of $\angle ECD$ .

The legs of a stool make an angle of $35^o$ with the floor as shown in
Fig. Find the angle $x$ and $y$

Three lines $AB,\ CD$ and $EF$ intersect each other at $O$ . If $\angle AOE=30^o$ and $\angle DOB=40^o$ Fig. find $\angle COF$ .

In Fig., $AB \parallel CD$ . Find the reflex $\angle EFG$ .

In Fig. , If $RP =RQ$ , find the value of $x$

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Find the value of $\angle a$ .

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find the value of $x$ .

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find $\angle BPD$ .

Two straight lines AB and CD intersect each other at a point O and angle $AOC = 50^\circ,$ find:
angle BOD.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ then find $\angle BPC$ .

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 $\angle 4$

In the given figure, lines $AB$ and $CD$ intersect at point $O$ .
Name all the reflex angles formed and write the measures of each.

The adjoining figure, shows two straight lines AB and CD intersecting at P. If $\angle BPC = 4x - 5^\circ$ and $\angle APD = 3x + 15^\circ,$ find $\angle APD$ .

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.
$\angle 1$ and $\angle 10$

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.
$\angle 4$ and $\angle 5$

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.
$\angle 6$ and $\angle 12$

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:

$\angle 4 \quad and \quad \angle 6$

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:

$\angle 7 \quad and \quad \angle 13$

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:

$\angle 1 \quad and \quad \angle 8$

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:

$\angle 5 \quad and \quad \angle 7$

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:

$\angle 6 \quad and \quad \angle 8$

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:

$\angle 10 \quad and \quad \angle 12$

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:

$\angle 3 \quad and \quad \angle 6$

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:

$\angle 1 \quad and \quad \angle 5$

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:

$\angle 4 \quad and \quad \angle 6$

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:

$\angle 2 \quad and \quad \angle 13$

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:

$\angle 4 \quad and \quad \angle 5$

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:

$\angle 7 \quad and \quad \angle 9$

In the above figure, lines $AB$ and $CD$ intersect at $O$ . If $\angle AOC + \angle BOE = 70^{\circ} and \angle BOD = 40^{\circ}$ , find $\angle BOE$ and reflex $\angle COE$

Find the magnitude of $\angle BOD$

In figure, BA || ED and BC || EF. Show that $\angle ABC + \angle DEF = 180^o$

Find the measure of $x$ in the given figure.

In figure, HOPE is a parallelogram , Find the angle measure $x , y$ and $z$ . State the properties used to find them.

In Fig. arms BA and BC of $\angle ABC$ are respectively parallel to arms ED and EF of $\angle DEF$ . Prove that $\angle ABC= \angle DEF$ .

In Fig., line $n$ is a transversal to lines $l$ and $m$ . Identify the following:
Angle alternate to $\angle PQR$ and angle corresponding to angles $\angle RQF$ and angle alternate to $\ \angle PQE$ in Fig.