These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.

In Fig. 6.13 lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°. Find ∠BOE and reflex ∠COE.

Solution:

In this fig. we have given that

∠AOC + ∠BOE = 70° and ∠BOD = 40°

But, ∠BOD = ∠AOC (Vertical opposite angles)

∠AOC = 40 ……(i) (∵ ∠BOD = 40°)

Again, ∠AOC + ∠BOE = 70° (Given)

40 + ∠ZBOE = 70° (from equation (i))

∠BOE = 70 – 40 = 30°

Again, ∠AOC + ∠COE + ∠EOB = 180° (Because AB is a st. line)

40° + ∠COE + 30° = 180° (∵ ∠AOC = 40° and ∠EOB = 30°)

Therefore, ∠BOE = 30°, and Reflex ∠COE = 360° – 110° = 250°

Therefore, ∠BOE = 30°, and Reflex ∠COE = 250°.

Question 2.

In Fig. 6.14 lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution:

We have given

∠POY = 90°, and a : b = 2 : 3

or, \(\frac{a}{b}=\frac{2}{3}\)

∴ a = \(\frac {2}{3}\) b ……(i)

Again, ∠POY + a + b = 180° (Because OXY is a straight line)

90 + \(\frac {2}{3}\) b + b = 180°

or, \(\frac {2}{3}\) b + b = 90

b = \(\frac{90 \times 3}{5}\) = 54° ……(ii)

Again, c + b = 180° (because MN is a straight line)

c + 54° = 180° (from equation (ii))

Therefore, c = 126°

Question 3.

In Fig. 6.15, ∠PQR = ∠PRQ then prove that ∠PQS = ∠PRT.

Solution:

We have given

∠PQR = ∠PRQ

Again,

∠PQR + ∠PQS = 180° ……(i) (Linear pair)

∠PRQ = ∠PRT = 180° ……(ii) (Linear pair)

From equation (i) and (ii)

∠PQR + ∠PQS = ∠PRQ + ∠PRT

But, ∠PQR = ∠PRQ

Therefore, ∠PQS = ∠PRT

Question 4.

In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Solution:

We have given x + y = w + z

Again, x + y + w + z = 360

or, x + y + x + y = 360 (∵ x + y = w + z)

or, 2(x + y) = 360°

or, x + y = 180°

∠AOB = 180°

which is straight line angle.

∴ AOB is a straight line.

Question 5.

In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac {1}{2}\) (∠QOS – ∠POS)

Solution:

We have given,

Ray OR is perpendicular to line PQ

Therefore, ∠QOR = ∠POR = 90° ……(i)

Now, ∠QOS = ∠QOR + ∠ROS

or, ∠QOS = 90° + ∠ROS ……(ii) (from equation(i))

Again, ∠POS = ∠POR – ∠ROS

or, ∠PSO = 90 – ∠ROS ……(iii) (from equation (i))

Subtract equation (iii) from equation (ii),

∠QOS – ∠POS

∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)

∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS

∠QOS – ∠POS = 2∠ROS

∠ROS = \(\frac {1}{2}\) (∠QOS – ∠POS)

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:

We have given that ∠XYZ = 64°

and ∠ZYQ = ∠QYP (∵ YQ is bisector of ∠ZYP)

Now, ∠XYZ + ∠ZYQ + ∠QYP = 180° (because XYP is a straight line)

or, 64° + ∠ZYQ + ∠ZYQ = 180° (∵ ∠ZYQ + ∠QYP)

or, 2∠ZYQ = 180° – 64° = 116°

or, ∠ZYQ = 58°

Now, ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58°

∠XYQ = 122°

Again, ∠QYP = ∠ZYQ (from equation (i))

∴ ∠QYP = 58° (∵ ∠ZYQ = 58°)

∴ Reflex ∠QYP = 360° – 58° = 302°