SChand Composite Mathematics Class 7 Chapter 10 Lines and Angles Exercise 10A

 Exercise 10A

Question  1

Refer to figure. Find the value of x in all the following diagrams: 

(i) (DIAGRAM TO BE ADDED)

Sol: $x+60^{\circ}=360^{\circ}$

$x=360^{\circ}-60^{\circ}$

$x=300^{\circ}$ Ans

(ii) (DIAGRAM TO BE ADDED)

Sol:

 $\begin{aligned} x+70^{\circ}+50^{\circ} &+160^{\circ} \\ &=360^{\circ} \end{aligned}$

$x+280^{\circ}=360^{\circ}$

$x=360^{\circ}-280^{\circ}$

$x=80^{\circ} \quad$ Ans

(iii)  (DIAGRAM TO BE ADDED)

Sol: $x+30^{\circ}+40^{\circ}+140^{\circ}+90^{\circ}=360^{\circ}$

$x+300^{\circ}=360^{\circ}$

$x=360^{\circ}-300$

$x=60^{\circ}$ Ans 

Question  2

Refer to figure, $\angle A O C=100^{\circ}$, find $\angle B O C$.

Sol: (Diagram to be added)

AOB is  straight line 

= $\angle A O C$ and $\angle B O C$ form a linear pair 

= $\angle A O C+\angle B O C=180^{\circ}$

$100+\angle B O C=180^{\circ}$

= $\angle B O C=180^{\circ}-100 \Rightarrow \angle B O C=80^{\circ}$

Question  3

Determine the value of x in the figure.

(diagram to be added)

Sol: $\Rightarrow \angle P R S+\angle Q R S=180^{\circ}$

$\Rightarrow 4 x+6 x=180^{\circ}$

$\Rightarrow 10 x=180^{\circ} \Rightarrow x=\frac{180}{10}$

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

Question  4

Refer to figure. Copy and complete the table: 

Sol: (diagram to be added)

$\begin{array}{|c|c|c|}\hline \angle A B D & \angle D B E & \angle E B C \\\hline 70^{\circ} & 70^{\circ} & 40^{\circ} \\\hline 50^{\circ} & 70^{\circ} & 60 \\\hline 35^{\circ} & 80^{\circ} & 65 \\\hline\end{array}$

Question  5

Two lines intersect in a point as shown in the figure copy and complete the table beside the figure: 

Sol: (diagram to be added)

Question  6

In Q.6 and 7  l,m,n are straight lines . find the measure of each angle shown, without using a protractor. 

Sol: (IMAGE TO BE ADDED)

b = $122^{\circ}$ (vertical angles are equal)

a=c (Vertical angles)

$\Rightarrow a+b+c+122^{\circ}=360^{\circ}$

$\Rightarrow a+122^{\circ}+a+122^{\circ}=360^{\circ}$

$\Rightarrow 2 a+244=360^{\circ}$

$\Rightarrow 2 a=360^{\circ}-244$

$a=\frac{116}{2}$ 

$a=58^{\circ}$

$c=58^{\circ}$

Question  7

 (IMAGE TO BE ADDED)

Sol: $a=50^{\circ}$ $d=52^{\circ} \quad \Rightarrow c=b \quad$ (vertical oppasite angles)

$52^{\circ}+50^{\circ}+b=180^{\circ}$

$102^{\circ}+b=180^{\circ} \Rightarrow b=180^{\circ}-102$

$b=78^{\circ}$

$b=c=78^{\circ}$. (vertical angle)

Question  8

Find the angles in each of the following: 

(a) The angles are complementary and the smaller 40 less than the larger. 

Sol: One is x ; Other is y 

y= x-40 ; x+y =90 

=x$+x-40=90^{\circ}$ $\Rightarrow \quad 2 x=90^{\circ}+40^{\circ}$

$2 x=130^{\circ}$

$\Rightarrow x=\frac{130^{\circ}}{2} \quad \Rightarrow x=65^{\circ}$

$\Rightarrow y=65-40=25^{\circ}$

(b) The angles are complementary and the larger is four times the smaller .

Sol: One angle = x  ; Other = y 

y = 4x = $x_{1}+y=90$

$\Rightarrow x+4 x=90 \Rightarrow 5 x=90 \Rightarrow x=\frac{90}{5}$

$\Rightarrow x=18^{\circ} \quad \Rightarrow y=4 \times 18^{\circ} \quad \Rightarrow y=72^{\circ}$ Ans

(c) The angle are supplementary and the smaller is one half of the larger. 

Sol: $x ; y \quad \Rightarrow \quad x=\frac{y}{2}$

$\Rightarrow x+y=180^{\circ} \quad \Rightarrow \quad \frac{y}{2}+y=180^{\circ} \Rightarrow \frac{y+2 y}{2}=180^{\circ}$

$\Rightarrow \quad 3 y=360^{\circ}$

$\Rightarrow y=\frac{360^{\circ}}{3}$

$y=120^{\circ}$

$x=\frac{y}{2}=\frac{120^{\circ}}{2}=60^{\circ}$ answer 

(d) The angles are supplementary and the larger is $58^{\circ}$ more than the smaller .

Sol:  $\Rightarrow x+y=180^{\circ} \quad \Rightarrow \quad x+x+53^{\circ}=180^{\circ} \Rightarrow 2 x=180^{\circ}-58^{\circ}$

$\Rightarrow 2 x=122^{\circ} \quad \Rightarrow \quad 21=61^{\circ}$

$y=x+58^{\circ}=61+58=119^{\circ}$

(e) The angles are supplementary and the larger is 20 less than three times the smaller. 

Sol: $x: y \quad \Rightarrow y=3 x-20^{\circ}$

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

$\Rightarrow x+3 x-20^{\circ}=180^{\circ} \Rightarrow 4 x=180^{\circ}+20^{\circ}$

$\Rightarrow 4 x=200^{\circ} \Rightarrow x=\frac{200}{4} \Rightarrow x=50^{\circ}$

$=y=3 \times 50-20^{\circ}=150 - 20^{\circ}$

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

(f) The angles are adjacent and form an angle $140^{\circ}$ The smaller is $20^{\circ}$ Less than the larger. 

Sol: $x ; y ; \quad x=y+28^{\circ}$

$\Rightarrow x+y=140^{\circ}$

$\Rightarrow y+28^{\circ}+y=140^{\circ}$

$\Rightarrow 2 y=140^{\circ}-28^{\circ}$

$\Rightarrow 2 y=112^{\circ}$

$\Rightarrow y=\frac{112^{\circ}}{2} \Rightarrow y=56^{\circ}$

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

$\Rightarrow x=56^{\circ}+28^{\circ}$

$\Rightarrow x=84^{\circ}$

Question  9

Calculate the value of x: 

(a) (diagram to be added)

Sol:  Straight line 

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

$\Rightarrow 5 x=\frac{180^{\circ}}{5}$

$\Rightarrow 2 x=\frac{180}{36^{\circ}}$ Answer 

(b)  (diagram to be added)

Sol: Right angle 

$\Rightarrow 4 x+521=90^{\circ}$

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

$\Rightarrow x=\frac{90}{9}$

$\Rightarrow x=10^{\circ}$

(c)  (diagram to be added)

Sol: $x+2 x+3 x+4 x=360^{\circ}$

$10 x=360^{\circ}$

$x=36^{\circ}$

(ii) (diagram to be added)

Sol: $\Rightarrow 3 x-5^{\circ}+x+20+65^{\circ}=180^{\circ}$

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

$\Rightarrow 4 x=180^{\circ}-80^{\circ}$

$\Rightarrow 4 x=100^{\circ}$

$\Rightarrow x=\frac{100}{4}$

$\Rightarrow x=25^{\circ}$

(iii) (diagram to be added)

Sol: $\Rightarrow 2 x+2 x+x+60^{\circ}=360^{\circ}$

$\Rightarrow 5 x=360^{\circ}-60^{\circ}$

$\Rightarrow 5 x=300^{\circ}$

$\Rightarrow \quad x=\frac{300}{5}$

$\Rightarrow \quad x=60^{\circ}$ Ans 

Question  10

IN the given figure line AB and CD interest at O. 

$\angle B O C=36^{\circ}$, find $\angle x,\angle y, \angle 2$.

Sol: $\angle x=\angle B O C=36^{\circ}$ (Vertical opposite angle )

$\angle y=\angle z$

$\Rightarrow \quad \angle B O C+\angle x+\angle y+\angle z=360^{\circ}$

$\Rightarrow \quad 36^{\circ}+36^{\circ}+\angle y+\angle y=360^{\circ}$

$72^{\circ}+2 \angle y=360^{\circ}$

$2 \angle y=360^{\circ}-72^{\circ}$

$\angle y=\frac{288}{2}$ =  $\angle Y=144^{\circ} \quad \angle z=144^{\circ}$

Question  11

In the given figure PQ and RS are two lines Intersecting each other at Point OB. IF $\angle R O T=90^{\circ}$ Find the value of x, y and z

Sol: (IMAGE TO BE ADDED)

OP and OQ opposite rays 

POQ is s straight line 

= $\angle P O R+\angle R O T+\angle T O Q=180^{\circ}$

$\Rightarrow 2 x+90^{\circ}+x=180^{\circ} .$

$\Rightarrow \quad 3 x=130^{\circ}-90^{\circ}$

$\Rightarrow \quad 3 x=90^{\circ}$

$\Rightarrow x=\frac{90}{3}$

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

$\Rightarrow y=2 \times 21 \Rightarrow y=2 \times 30$

y= $60^{\circ}$

$z=90^{\circ}+21 \quad$ (vertical opp)

$z=90+30$

$z=120^{\circ}$ Ans

Question  12

In the given figure , if 

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

Sol: (IMAGE TO BE ADDED)

= given $\angle A O C+\angle B O D=266^{\circ}$

$\angle A O C=\angle B O D \quad$ (vertical)

$\angle A O C+\angle A O C=266^{\circ} \Rightarrow 2 \angle A O C=266^{\circ}$

$\Rightarrow \quad \angle A O C=133^{\circ}$ Ans $\quad \angle B O D=133^{\circ}$ Ans.

OA and OB are opposite rays ∴ AOB is a straight line 

$\Rightarrow \angle A O C+\angle C O B=180^{\circ}$

$\Rightarrow 133^{\circ}+\angle \angle O B=180^{\circ} \Rightarrow \angle C O B=180^{\circ}-133^{\circ}$

$\angle C O B=47^{\circ}$

$\angle A O D=47^{\circ}$

Question  13

In the given figure , if $\angle A O C+\angle B O C+\angle B O D=338^{\circ}$ Find the four angles 

(image to be added)

Sol: $\angle A O C+\angle B O C+\angle B O D+\angle A O D=360^{\circ}$

$338^{\circ}+\angle A O D=360^{\circ}$

$\angle A O D=360^{\circ}-338^{\circ}$

$\angle A O D=22^{\circ}=\angle B O C$

$\angle A O D+\angle B O D=180$ Answer