RD Sharma Solutions for Class 9 is designed according to the syllabus prescribed by CBSE. In geometry, we generally come across geometrical figures consisting of angles which posses certain relations among themselves. In this exercise, students will study the relations between the angles formed between lines. Find detailed RD Sharma solutions chapter 8 exercise 8.2 – lines and angles below.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2

### Access Answers to Maths RD Sharma 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 = 25 ^{0}, what is the value of y?**

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

** **

**Solution:**

**(i)** Given: x = 25

From figure: âˆ AOC and âˆ BOC form a linear pair

Which implies, âˆ AOC + âˆ BOC = 180^{0}

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

âˆ AOC + âˆ 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}_{ .}Answer!!

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

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

(2y + 5) + 3x = 180

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

75 + 3x = 180

3x = 105

x = 35

Therefore, x = 35^{0}

**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 :

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

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

[As âˆ AOD + âˆ BOD = 180^{0}and âˆ AOC+ âˆ BOC = 180

^{0}.]

**Question 3 : In the given figure, find x. Further find âˆ BOC , âˆ COD and âˆ AOD.**

**Solution:**

From figure, âˆ AOD and âˆ BOD form a linear pair,

Therefore, âˆ AOD+ âˆ BOD = 180^{0}

Also, âˆ AOD + âˆ BOC + âˆ COD = 180^{0}

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

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

3x + 30 = 180

3x = 180 â€“ 30

x = 150/3

x = 50^{0}

Now,

âˆ AOD=(x+10) =50 + 10 = 60

âˆ COD = x = 50

âˆ BOC = (x+20) = 50 + 20 = 70

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

**Question 4: In figure, rays OA, OB, OC, OD and OE have the common end point 0. Show that âˆ AOB+âˆ BOC+âˆ COD+âˆ DOE+âˆ EOA=360Â°.**

**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:

âˆ AOB and âˆ BOX are linear pair angles, therefore,

âˆ AOB +âˆ BOX = 180^{0}

Or, âˆ AOB + âˆ BOC + âˆ COX = 180^{0} â€”â€“â€”â€“(1)

Also,

âˆ AOE and âˆ EOX are linear pair angles, therefore,

âˆ AOE+âˆ EOX =180Â°

Or, âˆ AOE + âˆ DOE + âˆ DOX = 180^{0} â€”â€“(2)

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

âˆ AOB + âˆ BOC + âˆ COF + âˆ AOE + âˆ DOF + âˆ DOE = 180^{0} + 180^{0}

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

Hence Proved.

**Question 5 : In figure, âˆ AOC and âˆ BOC form a linear pair. If a â€“ 2b = 30Â°, find a and b?**

**Solution:**

Given : âˆ AOC and âˆ BOC form a linear pair.

=> a + b = 180** ^{0 }** â€¦..(1)

a â€“ 2b = 30^{0} â€¦(2)** **(given)

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

a + b â€“ a + 2b = 180 â€“ 30

3b = 150

b = 150/3

b = 50^{0}

Since, a â€“ 2b = 30^{0}

a â€“ 2(50) = 30

a = 30 + 100

a = 130^{0}

Therefore, the values of a and b are 130Â° and 50Â° 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 :

(âˆ AOD,âˆ DOB),(âˆ DOB,âˆ BOC),(âˆ COA, âˆ AOD) and (âˆ BOC,âˆ 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 :

âˆ EOC and âˆ DOC

âˆ EOD and âˆ DOB

âˆ DOC and âˆ COB

âˆ EOD and âˆ DOA

âˆ DOC and âˆ COA

âˆ BOC and âˆ BOA

âˆ BOA and âˆ BOD

âˆ BOA and âˆ BOE

âˆ EOC and âˆ COA

âˆ EOC and âˆ 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 360Â°.

Therefore,

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

7x = 360^{0} â€“ 150^{0}

7x = 210^{0}

x = 210/7

x = 30^{0}

Hence, the value of x is 30Â°.

**Question 9: In figure, AOC is a line, find x.**

**Solution:**

From the figure, âˆ AOB and âˆ BOC are linear pairs,

âˆ AOB +âˆ BOC =180Â°

70 + 2x = 180

2x = 180 â€“ 70

2x = 110

x = 110/2

x = 55

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

**Question 10: In figure, POS is a line, find x.**

**Solution:**

From figure, âˆ POQ and âˆ QOS are linear pairs.

Therefore,

âˆ POQ + âˆ QOS=180^{0}

âˆ POQ + âˆ QOR+âˆ 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}.

## Access other exercise solutions of Class 9 Maths Chapter 8 Lines and Angles

## RD Sharma Solutions for Class 9 Maths Chapter 8 Exercise 8.2

Class 9 Maths Chapter 8 Lines and Angles Exercise 8.2 is based on topics such as some angle relations and linear pair of angles. Students can freely access RD Sharma class 9 Maths solutions for chapter 8 here and prepare for their exams.