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

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,