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

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Question 1.

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

Solution:

It is given that

∠AOC +∠BOE = 70° ………..(1)

∠AOC = ∠BOD

⇒ ∠AOC = 40°

Putting in eqn. (1),

40° + ∠BOE = 70°

∠BOE = 70° – 40°

⇒ ∠BOE = 30°

∠AOE + ∠BOE = 180° (Linear pair)

∠AOC + ∠COE + ∠BOE = 180°

⇒ ∠AOC + ∠BOE + ∠COE = 1800

⇒ 70° +∠COE = 180°

⇒ ∠COE = 180° – 70°

⇒ ∠COE = 110°

Now, Reflex ∠COE = 360° – 110° = 250°

Question 2.

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

Solution:

Let a = 2 x and b = 3x

∠POX + ∠POY = 180° (Linear pair)

∠POX + 90° = 180°

∠POX = 180° – 90° = 90°

⇒ a + b = 90°

⇒ 2x + 3x = 90°

⇒ 5x = 90°

⇒ x = \(\frac {90°}{5}\) = 18°

∴ a = 2x = 2 x 18°= 36°

⇒ b = 3x = 3 x 18°= 54°

Now, ∠XON + ∠YON = 180° (Linear pair)

⇒ c + 54° = 180° (∠XOM = b and ∠XON = c)

⇒ c = 180° – 54°

⇒ c = 126°

Question 3.

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

Solution:

Given: ∠PQR = ∠PRQ

To Prove: ∠PQS = ∠PRT

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

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

But ∠PQR =∠PRQ

Putting value ∠PQR from eqn. (3) in eqn. (1)

∠PQS + ∠PRQ = 180°

Now from eqn. (2) and (4), we have

∠PQS + ∠PRQ = ∠PRQ + ∠PRT

∠PQS = ∠PRT

Question 4.

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

Solution:

Given: x + y = w + z

To Prove: AOB is a line.

Proof: ∠BOC + ∠AOC +∠BOD + ∠AOD = 360°

(Sum of all angles round a point is equal to 360°)

x + y + w + z = 360° ……..(1)

But x + y = w + z ……..(2)

From eqn. (1) and (2), we have

w + z + w + z = 360°

⇒ 2w + 2z = 360°

⇒2(w + z) = 360°

⇒ w + z = \(\frac{360°}{2}\)

⇒ w + z = 180°

w + z = 180° (Linear pair)

Hence,

∠BOD + ∠AOD = 180°

Therefore AOB is a line.

Question 5.

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

∠ROS = \(\frac{1}{2}\)(∠QOS – ∠POS).

Solution:

Given: Ray OR ⊥ PQ and ray OS is another ray lying between OP and OR.

To Prove: ∠ROS = (∠QOS – ∠POS)

Proof: Ray OR ⊥ PQ

∴ ∠POR = ∠ROQ = 90°

Now, ∠POS = ∠POR – ∠ROS

∠POS = 90° – ∠ROS ………..(1)

∠QOS = ∠QOR + ∠ROS

∠QOS = 90° + ∠ROS ……..(2)

Subtracting eqn (2) from eqn (1),

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

∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS

∠QOS – ∠POS = 2∠ROS

2 ∠ROS = ∠QOS – ∠POS

∠ROS = \(\frac{1}{2}\)(∠QOS – ∠POS)

Question 6.

It is given that ∠XYZ = 640 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:

∠XYZ = 64°

Ray YZ stands on PX.

∠XYZ + ∠PYZ = 180° (Linear pair)

64° + ∠PYZ = 180°

⇒ ∠PYZ = 180°- 64°

⇒ ∠PYZ = 116°

⇒ ∠QYZ = \(\frac{∠PYZ}{2}\)

(YQ is the bisector of ∠PYZ)

∠QYZ = \(\frac{116°}{2}\)

∠QYZ = 58°

∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58°

∠XYQ = 122°

∠QYP = ∠QYZ

(YQ is the bisector of ∠PYZ)

∠QYP = 58°

Reflex ∠QYP = 360° – 58°

Reflex ∠QYP = 302°