RD Sharma Solutions for Class 9 are designed according to the syllabus prescribed by the CBSE. In geometry, we generally come across geometrical figures consisting of angles which possess certain relations among themselves. In this exercise, students will study the relations between the angles formed between lines. Find the detailed RD Sharma Solutions Class 9 Chapter 8 Exercise 8.2 – Lines and Angles below.
RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 8 Lines and Angles Exercise 8.2 Page Number 8.13
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?
Solution:
(i) Given: x = 25
From figure: ∠AOC and ∠BOC form 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 below figure, write all pairs of adjacent angles and all the linear pairs.
Solution: 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.
Solution:
From figure, ∠AOD and ∠BOD form 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°.
Solution:
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 line AX.
From figure:
∠AOB and ∠BOX are linear pair angles, therefore,
∠AOB +∠BOX = 1800
Or, ∠AOB + ∠BOC + ∠COX = 1800 —–—–(1)
Also,
∠AOE and ∠EOX are linear pair angles, therefore,
∠AOE+∠EOX =180°
Or, ∠AOE + ∠DOE + ∠DOX = 1800 —–(2)
By adding equations, (1) and (2), we get;
∠AOB + ∠BOC + ∠COX + ∠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?
Solution:
Given : ∠AOC and ∠BOC form 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 of adjacent angles are formed when two lines intersect at a point?
Solution: Four pairs of adjacent angles are formed when two lines intersect 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 of adjacent angles, in all, can you name in figure given?
Solution: 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.
Solution:
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.
Solution:
From the figure, ∠AOB and ∠BOC are linear 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 a line, find x.
Solution:
From figure, ∠POQ and ∠QOS are linear 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.
RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2
Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2 is based on topics such as some angle relations and the linear pair of angles. Students can access RD Sharma Class 9 Maths Solutions for Chapter 8 here for free and prepare well for their exams.
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