The problems are solved in such a way that they help students understand the topics and improve their problem solving abilities. Practising lots of examples before solving the main exercise makes them familiar with the formulas and methods. The solutions are prepared by experts at BYJUâ€™S with the aim of helping students score well in the board exams. In the CBSE exam pattern each step carries marks, so the answers are prepared with explanations to make it easy for understanding. RS Aggarwal Solutions for Class 9 Maths Chapter 7 Lines and Angles Exercise 7B are provided here.

## RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7B Download PDF

## Access RS Aggarwal Solutions for Class 9 Chapter 7: Lines and Angles Exercise 7B

__Exercise 7(B)__

**1. In the adjoining figure, AOB is a straight line. Find the value of x.**

**Solution:**

From the figure we know that âˆ AOC and âˆ BOC are a linear pair of angles

So we get

âˆ AOC + âˆ BOC = 180^{o }

We know that

62^{o} + x^{o} = 180^{o}

On further calculation

x^{o} = 180^{o} – 62^{o}

By subtraction

x^{o} = 118^{o}

Therefore, the value of x is 118.

**2. In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find âˆ AOC and âˆ BOD.**

**Solution:**

From the figure we know that âˆ AOB is a straight line

So we get

âˆ AOB = 180^{o}

It can also be written as

âˆ AOC + âˆ COD + âˆ BOD = 180^{o}

By substituting the values

(3x â€“ 7) ^{o} + 55^{o} + (x + 20) ^{o} = 180^{o}

3x â€“ 7^{o} + 55^{o} + x + 20^{o} = 180^{o}

On further calculation

4x + 68^{o} = 180^{o}

4x = 112^{o}

By division

x = 28^{o}

By substituting the value of x we get

âˆ AOC = (3x â€“ 7) ^{o}

= 3(28^{o}) â€“ 7^{o}

On further calculation

= 84^{o} – 7^{o}

By subtraction

= 77^{o}

âˆ BOD = (x + 20) ^{o}

= (28 + 20) ^{o}

By addition

= 48^{o}

**3. In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find âˆ AOC, âˆ COD and âˆ BOD.**

**Solution:**

From the figure we know that âˆ BOD and âˆ AOD are a linear pair of angles

So we get

âˆ BOD + âˆ AOD = 180^{o}

It can also be written as

âˆ BOD + âˆ COD + âˆ COA = 180^{o}

By substituting the values

x^{o} + (2x â€“ 19) ^{o} + (3x + 7) ^{o} = 180^{o}

x + 2x â€“ 19^{o} + 3x + 7^{o} = 180^{o}

On further calculation

6x â€“ 12^{o} = 180^{o}

6x = 180^{o} + 12^{o}

So we get

6x = 192^{o}

By division

x = 32^{o}

By substituting the value of x we get

âˆ AOC = (3x + 7) ^{o}

= 3(32^{o}) + 7^{o}

On further calculation

= 96^{o} + 7^{o}

By addition

= 103^{o}

âˆ COD = (2x – 19) ^{o}

= (2(32^{o}) – 19) ^{o}

So we get

= (64 â€“ 19) ^{o}

By subtraction

= 45^{o}

âˆ BOD = x^{o} = 32^{o}

**4. In the adjoining figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.**

**Solution:**

From the figure it is given that

x: y: z = 5: 4: 6

We can also write it as

x + y + z = 5 + 4 + 6 = 15

It is given that XOY is a straight line

So we know that

x + y + z = 180^{o}

As we know the sum of ratio is 15 then we can write that the measure of x as 5

The sum of all the angles in a straight line is 180^{o}

So we get the measure of x as

x = (5/15) Ã— 180

On further calculation

x = 60

As we know the sum of ratio is 15 then we can write that the measure of y as 4

The sum of all the angles in a straight line is 180^{o}

So we get the measure of y as

y = (4/15) Ã— 180

On further calculation

y = 48

In order to find the value of z

We know that

x + y + z = 180^{o}

Substituting the values of x and y we get

60^{o} + 48^{o} + z = 180^{o}

On further calculation

z = 180^{o} – 60^{o} – 48^{o}

By subtraction we get

z = 72^{o}

Therefore, the values of x, y and z are 60^{o}, 48^{o} and 72^{o}.

**5. In the adjoining figure, what value of x will make AOB, a straight line?**

**Solution:**

We know that AOB will be a straight line only if the adjacent angles form a linear pair.

âˆ BOC + âˆ AOC = 180^{o}

By substituting the values we get

(4x â€“ 36) ^{o} + (3x + 20) ^{o} = 180^{o}

4x – 36^{o} + 3x + 20^{o} = 180^{o}

On further calculation we get

7x = 180^{o} – 20^{o} + 36^{o}

7x = 196^{o}

By division we get

x = 196/7

x = 28

Therefore, the value of x is 28.

**6. Two lines AB and CD intersect at O. If âˆ AOC = 50 ^{o}, find âˆ AOD, âˆ BOD and âˆ BOC.**

**Solution:**

From the figure we know that âˆ AOC and âˆ AOD form a linear pair.

It can also be written as

âˆ AOC + âˆ AOD = 180^{o}

By substituting the values

50^{o} + âˆ AOD = 180^{o}

âˆ AOD = 180^{o} – 50^{o}

By subtraction

âˆ AOD = 130^{o}

According to the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

So we get

âˆ AOD = âˆ BOC = 130^{o}

According to the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

So we get

âˆ AOC = âˆ BOD = 50^{o}

**7. In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.**

**Solution:**

From the figure we know that âˆ COE and âˆ DOF are vertically opposite angles

âˆ COE = âˆ DOF = âˆ z = 50^{o}

From the figure we know that âˆ BOD and âˆ COA are vertically opposite angles

âˆ BOD = âˆ COA = âˆ t = 90^{o}

We also know that âˆ COA and âˆ AOD form a linear pair

âˆ COA + âˆ AOD = 180^{o}

It can also be written as

âˆ COA + âˆ AOF + âˆ FOD = 180^{o}

By substituting values in the above equation we get

90^{o} + x^{o} + 50^{o} = 180^{o}

On further calculation we get

x^{o} + 140^{o} = 180^{o}

x^{o} = 180^{o} – 140^{o}

By subtraction

x^{o} = 40^{o}

From the figure we know that âˆ EOB and âˆ AOF are vertically opposite angles

âˆ EOB = âˆ AOF = x = y = 40

Therefore, the values of x, y, z and t are 40, 40, 50 and 90.

**8. In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Hence, find âˆ AOD, âˆ COE and âˆ AOE.**

**Solution:**

From the figure we know that âˆ COE and âˆ EOD form a linear pair

âˆ COE + âˆ EOD = 180^{o}

It can also be written as

âˆ COE + âˆ EOA + âˆ AOD = 180^{o}

By substituting values in the above equation we get

5x + âˆ EOA + 2x = 180^{o}

From the figure we know that âˆ EOA and âˆ BOF are vertically opposite angles

âˆ EOA = âˆ BOF

So we get

5x + âˆ BOF + 2x = 180^{o}

5x + 3x + 2x = 180^{o}

On further calculation

10x = 180^{o}

By division

x = 180/10 = 18

By substituting the value of x

âˆ AOD = 2x^{o}

So we get

âˆ AOD = 2 (18) ^{o} = 36^{o}

âˆ EOA = âˆ BOF = 3x^{o}

So we get

âˆ EOA = âˆ BOF = 3 (18) ^{o} = 54^{o}

âˆ COE = 5x^{o}

So we get

âˆ COE = 5 (18) ^{o} = 90^{o}

**9. Two adjacent angles on a straight line are in the ratio 5:4. Find the measure of each one of these angles.**

**Solution:**

Consider the two adjacent angles as 5x and 4x.

We know that the two adjacent angles form a linear pair

So it can be written as

5x + 4x = 180^{o}

On further calculation

9x = 180^{o}

By division

x = 180/90

x = 20^{o}

Substituting the value of x in two adjacent angles

5x = 5 (20) ^{o} = 100^{o}

4x = 4 (20) ^{o} = 80^{o}

Therefore, the measure of each one of these angles is 100^{o} and 80^{o}

**10. If two straight lines intersect each other in such a way that one of the angles formed measures 90 ^{o}, show that each of the remaining angles measures 90^{o}.**

**Solution:**

Consider two lines AB and CD intersecting at a point O with âˆ AOC = 90^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

âˆ AOC = âˆ BOD = 90^{o}

From the figure we also know that âˆ AOC and âˆ AOD form a linear pair

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

Substituting the values

90^{o} + âˆ AOD = 180^{o}

On further calculation

âˆ AOD = 180^{o} – 90^{o}

By subtraction

âˆ AOD = 90^{o}

From the figure we also know that âˆ BOC = âˆ AOD are vertically opposite angles

âˆ BOC = âˆ AOD = 90^{o}

Therefore, it is proved that each of the remaining angle is 90^{o}

**11. Two lines AB and CD intersect at a point O such that âˆ BOC + âˆ AOD = 280 ^{o}, as shown in the figure. Find all the four angles.**

**Solution:**

From the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

âˆ AOD = âˆ BOC

It is given that

âˆ BOC + âˆ AOD = 280^{o}

We know that âˆ AOD = âˆ BOC

So it can be written as

âˆ AOD + âˆ AOD = 280^{o}

On further calculation

2 âˆ AOD = 280^{o}

By division

âˆ AOD = 280/2

âˆ AOD = âˆ BOC = 140^{o}

From the figure we know that âˆ AOC and âˆ AOD form a linear pair

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

Substituting the values

âˆ AOC + 140^{o} = 180^{o}

On further calculation

âˆ AOC = 180^{o} – 140^{o}

By subtraction

âˆ AOC = 40^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertically opposite angles

âˆ AOC = âˆ BOD = 40^{o}

Therefore, âˆ AOC = 40^{o}, âˆ BOC = 140^{o}, âˆ AOD = 140^{o} and âˆ BOD = 40^{o}

**12. Two lines AB and CD intersect each other at a point O such that âˆ AOC: âˆ AOD = 5:7. Find all the angles.**

**Solution:**

Consider âˆ AOC as 5x and âˆ AOD as 7x

From the figure we know that âˆ AOC and âˆ AOD for linear pair of angles

So it can be written as

âˆ AOC + âˆ AOD = 180^{o}

By substituting the values

5x + 7x = 180^{o}

On further calculation

12x = 180^{o}

By division

x = 180/12

x = 15^{o}

By substituting the value of x

âˆ AOC = 5x

So we get

âˆ AOC = 5 (15^{o}) = 75^{o}

âˆ AOD = 7x

So we get

âˆ AOD = 7 (15^{o}) = 105^{o}

From the figure we know that âˆ AOC and âˆ BOD are vertical angles

So we get

âˆ AOC = âˆ BOD = 75^{o}

From the figure we know that âˆ AOD and âˆ BOC are vertical angles

So we get

âˆ AOD = âˆ BOC = 105^{o}

**13. In the given figure, three lines AB, CD and EF intersect at a point O such that âˆ AOE = 35 ^{o} and âˆ BOD = 40^{o}. Find the measure of âˆ AOC, âˆ BOF, âˆ COF and âˆ DOE.**

**Solution:**

It is given that âˆ BOD = 40^{o}

From the figure we know that âˆ BOD and âˆ AOC are vertically opposite angles

âˆ AOC = âˆ BOD = 40^{o}

It is given that âˆ AOE = 35^{o}

From the figure we know that âˆ BOF and âˆ AOE are vertically opposite angles

âˆ AOE = âˆ BOF = 35^{o}

From the figure we know that AOB is a straight line

So it can be written as

âˆ AOB = 180^{o}

We can write it as

âˆ AOE + âˆ EOD + âˆ BOD = 180^{o}

By substituting the values

35^{o} + âˆ EOD + 40^{o} = 180^{o}

On further calculation

âˆ EOD = 180^{o} – 35^{o} – 40^{o}

By subtraction

âˆ EOD = 105^{o}

From the figure we know that âˆ COF and âˆ EOD are vertically opposite angles

âˆ COF = âˆ EOD = 105^{o}

**14. In the given figure, the two lines AB and CD intersect at a point O such that âˆ BOC = 125 ^{o}. Find the values of x, y and z.**

**Solution:**

From the figure we know that âˆ AOC and âˆ BOC form a linear pair of angles

So it can be written as

âˆ AOC + âˆ BOC = 180^{o}

By substituting the values we get

x + 125^{o} = 180^{o}

On further calculation

x = 180^{o} – 125^{o}

By subtraction

x = 55^{o}

From the figure we know that âˆ AOD and âˆ BOC are vertically opposite angles

So we get

y = 125^{o}

From the figure we know that âˆ BOD and âˆ AOC are vertically opposite angles

So we get

z = 55^{o}

Therefore, the values of x, y and z are 55^{o}, 125^{o} and 55^{o}

**15. If two straight lines intersect each other than prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.**

**Solution:**

Given: Let us consider AB and CD as the two lines intersecting at a point O where OE is the ray bisecting âˆ BOD and OF is the ray bisecting âˆ AOC.

To Prove: âˆ AOF = âˆ COF

Proof: EF is a straight line passing through the point O where

are two opposite rays.

From the figure we know that âˆ AOF and âˆ BOE, âˆ COF and âˆ DOE are vertically opposite angles

So it can be written as

âˆ AOF = âˆ BOE and âˆ COF = âˆ DOE

It is given that âˆ BOE = âˆ DOE

So we can write it as âˆ AOF = âˆ COF

Therefore, it is proved that âˆ AOF = âˆ COF.

**16. Prove that the bisectors of two adjacent supplementary angles include a right angle.**

**Solution:**

Given:

is the bisector of âˆ ACD and

is the bisector of âˆ BCD

To Prove: âˆ ECF = 90^{o}

Proof:

From the figure we know that

âˆ ACD and âˆ BCD form a linear pair of angles

So we can write it as

âˆ ACD + âˆ BCD = 180^{o}

We can also write it as

âˆ ACE + âˆ ECD + âˆ DCF + âˆ FCB = 180^{o}

From the figure we also know that

âˆ ACE = âˆ ECD and âˆ DCF = âˆ FCB

So it can be written as

âˆ ECD + âˆ ECD + âˆ DCF + âˆ DCF = 180^{o}

On further calculation we get

2 âˆ ECD + 2 âˆ DCF = 180^{o}

Taking out 2 as common we get

2 (âˆ ECD + âˆ DCF) = 180^{o}

By division we get

(âˆ ECD + âˆ DCF) = 180/2

âˆ ECD + âˆ DCF = 90^{o}

Therefore, it is proved that âˆ ECF = 90^{o}

### Access other exercise solutions of Class 9 Maths Chapter 7: Lines and Angles

Exercise 7A Solutions 12 Questions

Exercise 7C Solutions 24 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 7 – Lines and Angles Exercise 7B

RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles Exercise 7B is the second exercise which has solutions on finding adjacent angles, linear pair of angles and vertically opposite angles.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 7: Lines and Angles Exercise 7B

- Here the students mainly learn about angles, corollaries and theorems based on Lines and Angles.
- Revising the concepts of the chapter 7 is made easy with the help of solutions prepared by our faculty.
- PDF format of solutions can be downloaded by the students for free which is a vital resource to perform well in the exams.
- The solutions are prepared keeping the mind the CBSE exam pattern and weightage for the concepts.