RD Chapter 8- Lines and Angles Ex-8.1 |
RD Chapter 8- Lines and Angles Ex-8.3 |
RD Chapter 8- Lines and Angles Ex-8.4 |
RD Chapter 8- Lines and Angles Ex-VSAQS |

In the below Fig.OA and OB are opposite rays:

(i) If x = 25^{0},what is the value of y?

(ii) If y = 35^{0},what is the value of x?

**Answer
1** :

(i) Given:x = 25

From figure: ∠AOC and∠BOCform a linear pair

Which implies, ∠AOC + ∠BOC = 180^{0}

From the figure, ∠AOC = 2y + 5 and ∠BOC = 3x

∠AOC + ∠BOC = 180^{0}

(2y + 5) + 3x = 180

(2y + 5) + 3 (25) = 180

2y + 5 + 75 = 180

2y + 80 = 180

2y = 100

y = 100/2 = 50

Therefore, y = 50^{0}_{ }

(ii) Given:y = 35^{0}

From figure: ∠AOC + ∠BOC = 180° (Linear pair angles)

(2y + 5) + 3x = 180

(2(35) + 5) + 3x = 180

75 + 3x = 180

3x = 105

x = 35

Therefore, x = 35^{0}

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

**Answer
2** :

From figure, pairs of adjacent angles are :

(∠AOC, ∠COB) ; (∠AOD, ∠BOD) ;(∠AOD, ∠COD) ; (∠BOC, ∠COD)

And Linear pair of angles are (∠AOD, ∠BOD)and (∠AOC, ∠BOC).

[As ∠AOD + ∠BOD = 180^{0} and ∠AOC+ ∠BOC = 180^{0}.]

**Answer
3** :

From figure, ∠AOD and∠BODform a linear pair,

Therefore, ∠AOD+ ∠BOD = 180^{0}

Also, ∠AOD + ∠BOC + ∠COD = 180^{0}

Given: ∠AOD =(x+10)^{ 0} , ∠COD = x^{0} and ∠BOC = (x + 20)^{ 0}

( x + 10 ) + x + ( x + 20 ) = 180

3x + 30 = 180

3x = 180 – 30

x = 150/3

x = 50^{0}

Now,

∠AOD=(x+10)=50 + 10 = 60

∠COD = x= 50

∠BOC =(x+20) = 50 + 20 = 70

Hence, ∠AOD=60^{0},∠COD=50^{0} and∠BOC=70^{0}

In figure, rays OA,OB, OC, OD and OE have the common end point 0. Show that ∠AOB+∠BOC+∠COD+∠DOE+∠EOA=360°.

**Answer
4** :

Given: Rays OA, OB, OC, OD and OE have the common endpoint O.

Draw an opposite ray OX to ray OA, which make a straight lineAX.

From figure:

∠AOB and∠BOX arelinear pair angles, therefore,

∠AOB +∠BOX = 180^{0}

Or, ∠AOB + ∠BOC + ∠COX = 180^{0} —–—–(1)

Also,

∠AOE and∠EOX arelinear pair angles, therefore,

∠AOE+∠EOX =180°

Or, ∠AOE + ∠DOE + ∠DOX = 180^{0} —–(2)

By adding equations, (1) and (2), we get;

∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOE + ∠DOX = 180^{0} + 180^{0}

∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360^{0}

Hence Proved.

**Answer
5** :

Given : ∠AOC and∠BOCform a linear pair.

=> a + b = 180^{0 }…..(1)

a – 2b = 30^{0} …(2) (given)

On subtracting equation (2) from (1), we get

a + b – a + 2b = 180 – 30

3b = 150

b = 150/3

b = 50^{0}

Since, a – 2b = 30^{0}

a – 2(50) = 30

a = 30 + 100

a = 130^{0}

Therefore, the values of a and b are 130° and 50° respectively.

How many pairs ofadjacent angles are formed when two lines intersect at a point?

**Answer
6** :

Four pairs of adjacent angles are formed when two linesintersect each other at a single point.

For example, Let two lines AB and CD intersect at point O.

The 4 pair of adjacent angles are :

(∠AOD,∠DOB),(∠DOB,∠BOC),(∠COA, ∠AOD) and (∠BOC,∠COA).

How many pairs ofadjacent angles, in all, can you name in figure given?

**Answer
7** :

Number of Pairs of adjacent angles, from the figure, are :

∠EOC and∠DOC

∠EOD and∠DOB

∠DOC and∠COB

∠EOD and∠DOA

∠DOC and∠COA

∠BOC and∠BOA

∠BOA and∠BOD

∠BOA and∠BOE

∠EOC and∠COA

∠EOC and∠COB

Hence, there are 10 pairs of adjacent angles.

In figure,determine the value of x.

**Answer
8** :

The sum of all the angles around a point O is equal to 360°.

Therefore,

3x + 3x + 150 + x = 360^{0}

7x = 360^{0} – 150^{0}

7x = 210^{0}

x = 210/7

x = 30^{0}

Hence, the value of x is 30°.

**Answer
9** :

From the figure, ∠AOB and ∠BOC arelinear pairs,

∠AOB +∠BOC =180°

70 + 2x = 180

2x = 180 – 70

2x = 110

x = 110/2

x = 55

Therefore, the value of x is 55^{0}.

In figure, POS is aline, find x.

**Answer
10** :

From figure, ∠POQ and∠QOS arelinear pairs.

Therefore,

∠POQ + ∠QOS=180^{0}

∠POQ + ∠QOR+∠SOR=180^{0}

60^{0} + 4x +40^{0} = 180^{0}

4x = 180^{0} -100^{0}

4x = 80^{0}

x = 20^{0}

Hence, the value of x is 20^{0}.

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