RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions (2024)

In geometry, we frequently come across geometrical patterns consisting of angles that possess specific relations amid themselves. Here in the RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions, we will discuss the relations between the angles formed between lines, which is concerned under the Lines and Angles. Moreover, students will also learn about topics like- the relation of angles and linear pairs of angles.

For practice on these types of questions, we have provided the RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions PDF from which learners will get to solve several types of questions related to the angle relations and linear pair of angles. The PDF contains the questions from the 9th Class CBSE Text Book and the RD Sharma.

Download RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions PDF

Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2

Important Questions RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions

Linear Pair of Angle

When the angle connecting the two lines is 180 degrees (180°), they make a straight angle. A straight angle is simply another way to denote a straight line. A straight line can be envisioned as a circle with an endless radius. A line segment is any part of a line that has two endpoints. Also, a division of any line with only one endpoint is known as a ray. A line segment with P and Q as two endpoints is denoted as.

Adjacent Angles

Adjacent Angles share a standard vertex, a usual side, and no common interior points of angles. The Adjacent Angles share a vertex and side, but they don’t overlap.

Access the RD Sharma Chapter 8 Class 9 Maths Exercise 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?

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 the 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.

Fill in the Blanks Related to RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions

1. Suppose two parallel lines intersected by a transversal, then each pair of corresponding angles are_______.
2. Two lines perpendicular to the same line are______ to each other.
3. If a transversal meets a pair of lines in such a way that the sum of interior angles on the seine side of the transversal is 180°, then the lines are_______.
4. Two lines parallel to the similar line are________ to each other.

Answers to the above fill in the blanks-

1. Equal
2. Supplementary
3. Parallel
4. Parallel

FAQs Related to RD Sharma Chapter 8 Class 9 Maths Exercise 8.2 Solutions