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 $$\angle ABP+\angle CDP= \angle DPB$$.
1669423_38092efc1cda4042be58a75c6cdaa9e7.png


1656353_1669423_ans_d494328ee47441ada5bdfe8e31b641da.jpg


In figure, write another name for $$\angle 2$$.
1675452_788f6042aac0444290f7e54dff277b88.png

$$\angle 2$$ is nothing but $$\angle BOC.$$


Find $$x$$ and $$y$$.
1086671_44c80c0e7dfc4ca6baf5bf30b175ef97.png

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 

1088603_aeee34da31f3415d9fa69ad5420166dc.png

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 $$x : y : z = 2 : 3 : 5$$
570650.jpg

$$x:y:z=2:3:5$$
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 $$AB\parallel CD$$.
463653.jpg

$$x + 50^{\circ} = 180^{\circ}$$ (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 _______.
710482_4f68dcec724f43c5bd9d4f43018040b9.jpg

$$\angle YVZ$$.


Write all pairs of vertically opposite angles.
710517_5a5588d7e13c475dbe45fdc4ec3eaa5d.jpg

$$\angle XMQ$$ and $$\angle PMY, \angle XMP$$ and $$\angle YMQ$$.


Vertically opposite angle of $$\angle XVZ$$ is __________.
710481_ddfaf514e84d45f2a99a4aba4cf4d04f.jpg

Vertically opposite angle of given angle is $$\angle WVY$$.


Find x:
1029103_d83732a49408402ba95254a155d55c8a.png

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?
711120_3f8c90472c844740979ab6b4006ab2fb.png

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.
711123_8727aba17e8f4addbc8ac9212c0a9d65.png

$$\angle ACY$$ 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) $$\angle DEF$$
1057994_dacc943ff3004ea887bf256db2a6a5f9.png

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.
1159270_638e1610989d467b87e6e5245d8e8cc8.png

If $$p \parallel q$$ 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). 

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

1072779_1159270_ans_0a8b7fc850b54b28a915a27b496e7efc.png


In the given figure, if $$\angle A O C + \angle B O D = 266 ^ { \circ }$$ find all the four angles.
1228417_7344d91ba38d4df7a5060d15294f461b.png

Given $$\angle AOC+\angle BOD=266°$$

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

so $$\angle AOC+\angle BOD=266°$$

$$\Rightarrow 2\angle AOC=266°$$

$$ \Rightarrow \angle AOC=133°$$

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

so  $$133°+\angle AOD=180°$$

$$ \Rightarrow \angle AOD=180°-133°$$

$$\Rightarrow \angle AOD=47°$$

$$\angle BOC=47°$$   (opposite angle of $$\angle AOD$$)

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


In the given figure, find $$x$$.
1503270_6a882f094d484769b0308a4140413623.png

REF.Image
Opposite angle are equal
$$\therefore x=30^{\circ}$$
Now, $$30+30+y=180^{\circ}$$
$$y=120^{\circ}$$

1228123_1503270_ans_956015147927404199a382fa015f558f.png


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 $$c$$.
1791098_85808c802ccf433c92875ac98693a512.png

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,
$$\Rightarrow \angle a=30^o$$
$$\Rightarrow \angle b=150^o$$
$$\Rightarrow \angle c=150^o$$

1669977_1791098_ans_1643c184a73343648bcecb3ea0802bff.png


If two angles have a common vertex and their arms form opposite rays Fig., Then, how many types of angles are formed?
1791077_71092305862d4746a799a11a07f9430e.png

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.$$\quad \quad [Hence\ proved]$$

948727_243953_ans_2efb94874ac1426c8e63c8125aaecb26.png


Name all the pairs of vertically opposite angles in the following figure.
1073868_340446eb70d8478c987dd7d57b69e4e8.png

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 $$\angle SQT$$.
463683.jpg

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 $$CD$$

952308_558686_ans_a1b7cdb3e2cc41d892dd9d7ce9ca0023.PNG


In the Fig.  lines PQ and RS intersect each other at point O. If $$\angle POR : \angle ROQ = 5 : 7$$, Find all the angles.
570644.jpg

$$\angle POR  + \angle ROQ = 180^o$$    (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.

1017862_1026768_ans_781e6917e913471586e3dc83bd9e319b.png


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 $$\angle BOD$$.
1018344_3fada6e0ebc74aeea759f7e36e1b7c94.png

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 $$\angle PRT\, = \,40^\circ \,,\,\,\angle RPT\, = \,95^\circ \,\,and\,\,\angle TSQ\, = \,75^\circ \,,\,find\,\,\angle SQT$$.
1056303_e4cae7c0d17c459d8cfab2f9c8f0c0d9.png

$$In\,RPT$$

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

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

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

$$45$$

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

$$STQ = 45$$

$$Now\,in\,\,STQ$$

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

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

$$180 - 120$$

$$\angle SQT = 60$$



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 $$\angle EOC$$.
1043127_4919a54f40654814914b761dd9490a71.png

$$\angle BOE+\angle AOE=180^{\circ}$$ (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
1077241_bf58859d244543fab04dfbe4824eff43.png

 

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.
1148185_80a2cff2552743f2af5c0a6586be49e6.png


$$\textbf{Step -1: Given figure}$$
                  $$\text{Given figure is,}$$
               
                $$\text{Here }43^{\circ}\text{ and }b\text{ are the vertically opposite angles of the given intersecting straight lines.}$$
$$\textbf{Step -2: Apply property of angles}$$
                  $$\text{So, they are equal}\textbf{ (because vertically opposite angles are equal )}$$
                  $$\therefore{b}=43^{\circ}$$
$$\textbf{Hence the value of b is }\mathbf{43^{\circ}.}$$



In figure the value of $$x$$, hence find all three indicated angles.
1148645_4dd25f47fd344ee281595ef00538e5bb.png

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 $$\dfrac{x+u+y+z}{50}$$
1132588_4e447d6a6e8e4eddb05ea22a4876c03e.png

$$x=z\\u=y..............(1)$$
 (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.
1173314_1ba24872102c41ff975c6480fbdb4c2d.png

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 
$$ 3x+2 = 62 $$
$$ \Rightarrow 3x = 60 $$
$$ \Rightarrow x = 20 $$ 

1104074_1173314_ans_2877878e1aa94d32af8cf8b6c333a77d.png


Observe the following figure and write each:
(i) Linear pair
(ii) Pair of vertically opposite angles
1338012_b7b71e4d306e41d1a0e7cb27364648fd.png

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.
1345738_be7434f719f64f88a1d2afdf9831d97c.png

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 $$\angle{COE}$$
1405184_b892285908384fc3a6aaa89792dc1f53.png

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.

1669317_55185cf95bd24454851a5414f13a97cf.png


1656321_1669317_ans_e00f8041ae9b401ea51b9f5da9ea423e.jpg


In Fig., PQ and RS intersect each other at point O. If $$ \angle POR : \angle ROQ = 5:7,$$ find all the angles.
1669412_b9cd136063244ac9ac4e22af5cbedaa6.png

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 $$\angle 3$$.
1675453_a4a2414cc13c46a395086db8e0575b23.png

$$\angle 3$$ 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 $$\angle 4$$.
1677467_c231dba62359467bb1bc71550fb1fc12.png

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.
1677477_817ec3ffebfb480e834a7448dec868ec.png

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 $$u$$.
1677502_b6fa464914364ebca2e853e959452557.png

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 $$\angle DEF$$.
1677603_b65ab50ff4b243cd8983a908d7f3f8e1.png

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 $$u$$.
1677504_03ac69d3c0344c26915bdbec7bba646a.png

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 $$z$$.
1677505_a12d9137fc334b3c9f1dc8884f3ecf8e.png

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 $$\angle BOC$$
1715094_e774bcd3118b4081b8f116a1216c2dce.PNG

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 $$t$$
1715099_f8889f318871472e8d359d94a6f50963.png

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.
1715226_c988242c563f48a8acb73a6c373e5d28.png

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 $${90}^{o}$$.
1715109_87e53d2eb3884e56a044f7fc9495c5bf.png

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 $$z$$
1715236_19ce7961bd3b43dc98c562e7b858f862.png

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.
1821865_3eb926eaa0f24fc0a4a93f1645366c54.png

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.
1821867_2021e33d77c84ab7898b45d59f3388b2.png

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 $$\angle DOE$$
1715232_d2f5aaac62564114bd103faffa96c1a0.png

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 $$ \angle ECD $$.
1780988_2b3ab3736a774b80b577511521d8e69f.png

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 $$y$$
1791101_931bfc148018423a8efa4860f3179c4c.png

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 $$\angle COF$$.
1791148_7e0762d4e0a744cbb8a2709e12e90e81.png

As $$AB$$ 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 $$\angle EFG$$.
1791218_a8f9a3ea84664b7bafd22218d0bc97c7.png

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 $$x$$

1792779_5dd9adade600424089d5ebd8d99f7ab0.png

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.
1803202_02af69d1a87e4cbda8714587cfa10522.jpg

In $$\Delta$$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 $$\angle a$$.
1821862_df5cdb0c29e3424197e508c7cdeb2317.png

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 $$x$$.
1823225_db22cacacc1b47f5b5765085c284e4b5.png

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 $$\angle BPD$$.
1823228_5b28d2be8c38447d8ee23d0670cb07e2.png

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.

1823219_5dfff9c747cb4799a7e72d039848de32.png

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 $$\angle BPC$$.
1823230_1634f04567ae450f868b55dfcc2e3696.png

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 $$\angle 4$$
1823243_90aa5bc167a34cbcad88669c4f562a3d.png

$$\angle 2$$ 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.
1821872_eaba75c503134973bb35d0a1b909376e.png

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 $$\angle APD$$.
1823226_a6018556f7ae4177984108c25f315a52.png

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 $$\angle 10$$
1823271_7887d85c2adb46bfa93511b3d44f596f.png

$$\angle 1$$ 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 $$\angle 5$$
1823245_3ec4dd7b743543bb9bbb161586453740.png

$$\angle 4$$ 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 $$\angle 12$$
1823284_7cc41d0122094db2b643784d614b93d4.png

$$\angle 6$$ 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:

$$ \angle 4 \quad and \quad \angle 6 $$
1828416_8364dedaea30446387a7761d8522daf9.png

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

$$ \angle 7 \quad and \quad \angle 13 $$
1828498_4dc32e94734c4f69882710dd8f366190.png

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

$$ \angle 1 \quad and \quad \angle 8 $$
1828420_a9caecb2db644602a7e3b0cbdc259db3.png

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

$$ \angle 5 \quad and \quad \angle 7 $$
1828431_1ee36f6abbbe456681fdff8c2565b99a.png

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

$$ \angle 6 \quad and \quad \angle 8 $$
1828502_1cfc01e9764c4ee394cb55bd36a27d7f.png

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

$$ \angle 10 \quad and \quad \angle 12 $$
1828494_8229a613aa964359b919d2a5b631e1b1.png

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

$$ \angle 3 \quad and \quad \angle 6 $$
1828402_034668053be543d0be9aeab68875a88b.png

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

$$ \angle 1 \quad and \quad \angle 5 $$
1828424_786a180bb03d46d791460f4a98dd2649.png

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

$$ \angle 4 \quad and \quad \angle 6 $$
1828514_2f3c7f598a9941eba54411246af5fe9e.png

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

$$ \angle 2 \quad and \quad \angle 13 $$
1828520_00167855e061482b875bd500043202bb.png

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

$$ \angle 4 \quad and \quad \angle 5 $$
1828510_1e69e6a2ce664fe9b62caaa9f4b9d214.png

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

$$ \angle 7 \quad and \quad \angle 9 $$
1828508_a9be37ca9f6542e383088293c6a2271a.png

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


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

1828677_f114af10f7aa41c898c81880c73a2854.png

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 $$\angle COE$$
1869591_907a84a5f52b4de486b6bb2258f09280.PNG

$$\angle AOC = \angle BOD = 40^{\circ}$$.......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.
1879592_690132b7341d409d81b567bd0b63eef7.png

$$\because$$ $$AB$$ 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 $$\angle ABC + \angle DEF = 180^o$$
1879514_99dcc8e686744836b34523d9877e75d5.png

$$\because BA || ED$$
$$\angle 1 = \angle 2$$.......(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 $$ \angle \mathrm{SQT} $$
1879525_1e78ad1469f144208fba1f488fd3b56e.png

$$ \ln \Delta \mathrm{PRT} $$
$$ \angle \mathrm{RPT}+\angle \mathrm{PRT}+\angle \mathrm{RTP}=180^{\circ} \quad$$(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.
1346719_ebd89f71f41143589efb8e8257a0d7a4.png

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

                                                   

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

                $$\implies \angle D=70$$

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

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

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

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

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



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

1362041_53c9b5d65d804894a0a70dc651fe186e.png


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 $$\angle ABC= \angle DEF$$.
1669449_4933dbf498ac405eb6183337060a3fc5.png


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.

1675014_33c3b51acf8041d681433bbe91639bab.png

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 $$∠ARB$$.


Class 7 Maths Extra Questions