Exercise 6.1

Q1. In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Given: Lines AB and CD intersect at O.  

To find: ∠BOE and reflex ∠COE

Solution: ∠AOC = ∠BOD = 40° (Vertically opposite angles)

But ∠AOC + ∠BOE = 70° (Given)

∴ ∠BOE = 70° – 40°

= 30° Ans.

Now, ∠AOC + ∠COE + ∠BOE = 180° (AB is a straight line)

∴ 40° + ∠COE + 30° = 180°

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180° – 70°

⇒ ∠COE = 110°

∴ Reflex ∠COE = 360° – 110° = 250° Ans.

Q2. In the given figure, lines XY and MN intersect at O. If ∠POY = 90° and a:b = 2:3, find c.

Given: Lines XY and MN intersect at O. ∠POY = 90° and a:b = 2:3.

To find: c

Solution: Let a = 2x and b = 3x

∠POY = 90° (Given)

∴ ∠POX = 90° (Linear pair)

⇒ a + b = 90°

⇒ 2x + 3x = 90°

⇒ 5x = 90°

⇒ x = 18°

∴ a = 2x = 36° and b = 3x = 54°

Now, b + c = 180° (Linear pair)

⇒ 54° + c = 180°

⇒ c = 180° – 54°

⇒ c = 126° Ans.

Q3. In the given figure, if ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Given: ∠PQR = ∠PRQ

To prove: ∠PQS = ∠PRT

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

∠PRT + ∠PRQ = 180° (2) (Linear pair)

From (1) and (2), we get

∠PQS + ∠PQR = ∠PRT + ∠PRQ

But ∠PQR = ∠PRQ (Given)

∴ ∠PQS = ∠PRT 

Q4. In the given figure, if x + y = w + z, then prove that AOB is a line.

Given: x + y = w + z

To prove: AOB is a line

Proof: x + y + z + w = 360° (Complete angle)

But x + y = w + z (Given)

∴ x + y + x + y = 360°

⇒ 2x + 2y = 360°

⇒ 2 (x + y) = 360°

⇒ x + y = 180°

∴ AOB is a line

Q5. In the given 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 = ½ (∠QOS – ∠POS).

Given: POQ is a line. Ray OR line PQ. OS is another ray lying between rays OP and OR.

To prove: ∠ROS = ½ (∠QOS – ∠POS)

Proof: OR ⊥ PQ (Given)

∴∠POR = 90°

∠POS + ∠ROS = 90° (1)

and ∠POS + ∠QOS = 180° [Linear pair]

⇒ (∠POS + ∠QOS)/2 = 180°/2

⇒ ½ ∠POS + ½ ∠QOS = 90° (2)

From (1) and (2), we get

∠POS + ∠ROS = ½ ∠POS + ½ ∠QOS

⇒ ∠ROS = ½ ∠POS + ½ ∠QOS – ∠POS

⇒ ∠ROS = ½ ∠QOS – ½ ∠POS

⇒ ∠ROS = ½ (∠QOS – ∠POS)

Q6. 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.

Given: ∠XYZ = 64° and XY is produced to point P. Ray YQ bisects ∠ZYP.

To find: ∠XYQ and reflex ∠QYP

Solution: ∠XYZ + ∠ZYP = 180° (Linear pair)

∴ ∠ZYP = 180° – ∠XYZ = 180° – 64° = 116°

∠ZYQ = ½ ∠ZYP = ½ (116°) = 58°

∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122° Ans.

∠QYP = ∠ZYQ = 58°

∴ Reflex ∠QYP = 360° – 58° = 302° Ans.