Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.1 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.1

Question 1.

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

Solution:

Since OA and OB are opposite rays. Therefore, AB is a line.

Since ray OC stands on AB. Therefore,

∠AOC + ∠COB = 180° [Linear Pairs]

⇒ ∠AOC + ∠COE + ∠BOE = 180° [∵∠COB = ∠COE + ∠BOE]

⇒ (∠AOC + ∠BOE) + ∠COE = 180°

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

⇒ ∠COE = 180° – 70° = 110°

∴ Reflex ∠COE = 360° – 110° = 250°

Since OC and OD are opposite rays. Therefore, CD is a line.

Since ray OE stands on CD. Therefore,

∠COE + ∠EOD = 180° [Linear Pairs]

⇒ ∠COE) + ∠BOE + ∠BOD = 180°

⇒ 110° + ∠BOE + 40° = 180°

[∵ ∠COE = 110° (probed above), ∠BOD = 40° (Given)]

⇒ ∠BOE = 180° – 110° – 40° = 30°

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

Question 2.

In figure, lines XY and MN intersect at O. If ∠POY = 90° and a : 6 = 2 : 3, find c.

Solution:

Since a : b = 2 : 3 and a + b = ∠POX = ∠POY = 90° and sum- of ratios = 2 + 3 = 5

∴ a = \(\frac { 2 }{ 5 }\) x 90° = 2 x 18° = 36°

and b = \(\frac { 3 }{ 5 }\) x 90° = 3 x 18° = 54°

Since OM and ON are opposite rays. Therefore, MN is a line.

Since ray OX stands on MN. Therefore,

∠MOX + ∠XON = 180°’ [Linear Pairs]

⇒ c + b = 180° ⇒ c + 54° = 180°

⇒ c = 180° – 54° = 126°

Hence, c = 126°.

Question 3.

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

Solution:

Since QS and QR are opposite rays. Therefore, SR is a line.

Since QP stands on the line SR.

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

Again, RQ and RT are opposite rays. Therefore, QT is a line.

Since ER stands on the line QT.

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

From (1) and (2), we have

∠PQS + ∠PQR = ∠PRQ + ∠PRT [∵ Each side = 180°] … (3)

Also ∠PQR = ∠PRQ [Given] … (4)

Subtracting (4) from (3), we have ∠PQS = ∠PRT.

Question 4.

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

Solution:

Since the sum all angles round a point is equal to 360°

Thus, ∠BOC and ∠COA, ∠BOD and ∠AOD form linear pairs. Consequently OA and OB are two opposite rays. Therefore, AOB is a straight line.

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 ∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS).

Solution:

Since OR is perpendicular to the line PQ.

∴∠POR = ∠ROQ [∵ Each = 90°]

⇒ ∠POS + ∠ROS = ∠QOS – ∠ROS

⇒ 2∠ROS =∠QOS – ∠POS

⇒ ∠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:

Since XY is produced to point P. Therefore, XP is a straight line.

Since YZ stands on XP.

∴ ∠XYZ + ∠ZYP = 180° [Linear Pair]

⇒ 64° + ∠ZYP = 180° [∵ ∠XYZ = 64°]

⇒ ∠ZYP = 180° – 64° = 116°

Since ray YQ bisects ∠ZYP.

Therefore,∠QYP = ∠ZYQ = \(\frac { 111 }{ 2 }\) = 58°

Now, ∠XYQ = ∠XYZ + ∠ZYQ

⇒ ∠XYQ = 64° + 58° = 122°

and reflex ∠QYP = 360° – ∠QYP = 360° – 58° = 302°