NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Ex 6.1 Class 9 Maths** **Question 1

In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:

Since AB is a straight line,

∴ ∠AOC + ∠COE + ∠EOB = 180°

or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]

or ∠COE = 180° – 70° = 110°

∴ Reflex ∠COE = 360° – 110° = 250°

Also, AB and CD intersect at O.

∴∠COA = ∠BOD [Vertically opposite angles]

But ∠BOD = 40° [Given]

∴ ∠COA = 40°

Also, ∠AOC + ∠BOE = 70°

∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°

Thus, ∠BOE = 30° and reflex ∠COE = 250°.

Ex 6.1 Class 9 Maths** **Question 2.

In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Solution:

Since XOY is a straight line.

∴ b+a+∠POY= 180°

But ∠POY = 90° [Given]

∴ b + a = 180° – 90° = 90° …(i)

Also a : b = 2 : 3 ⇒ b = 3a/2 …(ii)

Now from (i) and (ii), we get

3a/2 + A = 90°

⇒ 5a/2 = 90°

⇒ a = 90∘/5×2=36∘ = 36°

From (ii), we get

b = 3/2 x 36° = 54°

Since XY and MN interstect at O,

∴ c = [a + ∠POY] [Vertically opposite angles]

or c = 36° + 90° = 126°

Thus, the required measure of c = 126°.

Ex 6.1 Class 9 Maths** **Question 3.

In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:

ST is a straight line.

∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]

Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]

From (1) and (2), we have

∠PQS + ∠PQR = ∠PRT + ∠PRQ

But ∠PQR = ∠PRQ [Given]

∴ ∠PQS = ∠PRT

Ex 6.1 Class 9 Maths** **Question 4.

In figure, if x + y = w + ⇒, then prove that AOB is a line.

Solution:

Sum of all the angles at a point = 360°

∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°

But (x + y) = (⇒ + w) [Given]

∴ (x + y) + (x + y) = 360° or,

2(x + y) = 360°

or, (x + y) = 360∘/2 = 180°

∴ AOB is a straight line.

Ex 6.1 Class 9 Maths** **Question 5.

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Solution:

rara POQ is a straight line. [Given]

∴ ∠POS + ∠ROS + ∠ROQ = 180°

But OR ⊥ PQ

∴ ∠ROQ = 90°

⇒ ∠POS + ∠ROS + 90° = 180°

⇒ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS … (1)

Now, we have ∠ROS + ∠ROQ = ∠QOS

⇒ ∠ROS + 90° = ∠QOS

⇒ ∠ROS = ∠QOS – 90° ……(2)

Adding (1) and (2), we have

2 ∠ROS = (∠QOS – ∠POS)

∴ ∠ROS = 1/2(∠QOS−∠POS)

Ex 6.1 Class 9 Maths** **Question 6.

It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.

Solution:

XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°

⇒ 64° + ∠ZYQ + ∠QYP = 180°

[∵ ∠XYZ = 64° (given)]

⇒ 64° + 2∠QYP = 180°

[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°

⇒ ∠QYP = 116∘/2 = 58°

∴ Reflex ∠QYP = 360° – 58° = 302°

Since ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]

⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]

Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

## Lines and Angles Class 9 Extra Questions Maths Chapter 6

**Extra Questions for Class 9 Maths Chapter 6 Lines and Angles**