Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Textbook Exercise Questions and Answers.

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

Question 1.

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

Answer:

Here, ∠AOC = ∠BOD (Vertically opposite angles)

Therefore, ∠AOC = 40° ………….(1)

(Since ∠BOD = 40°, given)

So, 40° + ∠BOE = 70° [From (1)]

Therefore, ∠BOE = 70° – 40° = 30° …………….(2)

Now, ∠AOC + ∠COB = 180° (Linear pair)

So, ∠AOC + ∠COE + ∠BOE = 180°

or 40° + ∠COE + 30° = 180° [From (1) and (2)1

So, ∠COE = 180° – 40° – 30° = 110°

Therefore, Reflex ∠COE = 360° – 110° = 250°

Question 2.

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

Answer:

Here, a : b = 2 : 3 and a + b = ∠POX = ∠POY = 90°

Let the angles be 2x and 3x

a = 2x, b = 3x

∠POX = a + b

90° = 2x + 3x

⇒ 5x = 90°

x = \(\frac{90^{\circ}}{5}\) = 18°

∴ a = 2x = 2 × 18° = 36°

b = 3x = 3 × 18° = 54°

Also, MN is a line.

Since ray OX stands on MN, therefore

∠MOX + ∠XON = 180°

or b + c = 180°

⇒ c = 180° – 54° = 126°

Hence, c = 126°

Question 3.

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

Answer:

SRT is a line.

Since QP stands on the line SRT, therefore

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

Since RP stands on the line SRT, therefore

∠PRQ + ∠PRT = 180° (Linear pair) ………… (2)

From (1) and (2), we have :

∠PQS + ∠PQR = ∠PRQ + ∠PRT …………(3)

Also, ∠PQR = ∠PRQ (Given) …………(4)

Subtracting (4) from (3), we have :

∠PQS = ∠PRT.

Hence Proved.

Question 4.

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

Answer:

Since the sum of all the angles round a point is equal to therefore

(∠BOC + ∠COA) + (∠BOD + ∠AOD) = 360°

⇒ (x + y) + (w + z) = 360°

But x + y = w + z (Given)

x + y = w + z = \(\frac{360^{\circ}}{2}\) = 180°

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

Hence proved.

Question 5.

In the 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).

Answer:

Since OR is perpendicular to the line PQ, therefore

∠POR = ∠ROQ (∵ Each = 90°)

or ∠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.

Answer:

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

Now ray YZ stands on XP.

∴ ∠XYZ + ∠ZYP = 180° (Linear pair)

⇒ 64° + ∠ZYP = 180° (∵ ∠XYZ = 64°)

∠ZYP = 180° – 64° = 116°

Since ray YQ bisects ZZYP,

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