RD Chapter 8 Lines and Angles Ex 8.2 Solutions
Question - 1 : - In the below Fig.OA and OB are opposite rays:
(i) If x = 250,what is the value of y?
(ii) If y = 350,what is the value of x?
Answer - 1 : -
(i) Given:x = 25
From figure: ∠AOC and∠BOCform a linear pair
Which implies, ∠AOC + ∠BOC = 1800
From the figure, ∠AOC = 2y + 5 and ∠BOC = 3x
∠AOC + ∠BOC = 1800
(2y + 5) + 3x = 180
(2y + 5) + 3 (25) = 180
2y + 5 + 75 = 180
2y + 80 = 180
2y = 100
y = 100/2 = 50
Therefore, y = 500
(ii) Given:y = 350
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 = 350
Question - 2 : - 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 = 1800 and ∠AOC+ ∠BOC = 1800.]
Question - 3 : - In the given figure, find x. Further find ∠BOC , ∠COD and ∠AOD.
Answer - 3 : -
From figure, ∠AOD and∠BODform a linear pair,
Therefore, ∠AOD+ ∠BOD = 1800
Also, ∠AOD + ∠BOC + ∠COD = 1800
Given: ∠AOD =(x+10) 0 , ∠COD = x0 and ∠BOC = (x + 20) 0
( x + 10 ) + x + ( x + 20 ) = 180
3x + 30 = 180
3x = 180 – 30
x = 150/3
x = 500
Now,
∠AOD=(x+10)=50 + 10 = 60
∠COD = x= 50
∠BOC =(x+20) = 50 + 20 = 70
Hence, ∠AOD=600,∠COD=500 and∠BOC=700
Question - 4 : - 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 = 1800
Or, ∠AOB + ∠BOC + ∠COX = 1800 —–—–(1)
Also,
∠AOE and∠EOX arelinear pair angles, therefore,
∠AOE+∠EOX =180°
Or, ∠AOE + ∠DOE + ∠DOX = 1800 —–(2)
By adding equations, (1) and (2), we get;
∠AOB + ∠BOC + ∠COF + ∠AOE + ∠DOE + ∠DOX = 1800 + 1800
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 3600
Hence Proved.
Question - 5 : - In figure, ∠AOC and ∠BOC form a linear pair. If a – 2b= 30°, find a and b?
Answer - 5 : -
Given : ∠AOC and∠BOCform a linear pair.
=> a + b = 1800 …..(1)
a – 2b = 300 …(2) (given)
On subtracting equation (2) from (1), we get
a + b – a + 2b = 180 – 30
3b = 150
b = 150/3
b = 500
Since, a – 2b = 300
a – 2(50) = 30
a = 30 + 100
a = 1300
Therefore, the values of a and b are 130° and 50° respectively.
Question - 6 : - 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).
Question - 7 : - 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.
Question - 8 : - 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 = 3600
7x = 3600 – 1500
7x = 2100
x = 210/7
x = 300
Hence, the value of x is 30°.
Question - 9 : - In figure, AOC is a line, find x.
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 550.
Question - 10 : - In figure, POS is aline, find x.
Answer - 10 : -
From figure, ∠POQ and∠QOS arelinear pairs.
Therefore,
∠POQ + ∠QOS=1800
∠POQ + ∠QOR+∠SOR=1800
600 + 4x +400 = 1800
4x = 1800 -1000
4x = 800
x = 200
Hence, the value of x is 200.