# Lines and Angle – Exercise 6.1 – Class IX

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

Data given :

Lines AB and CD intersect at O.

∠ AOC + ∠BOE = 70° …………….(1)

and ∠BOD = 40°…………………..(2)

Since ∠AOC = ∠BOD (vertically opposite angles)

Therefore, ∠AOC = 40° [FROM 2]

and 40° + ∠BOE = 70° [from 1]

∠BOE = 70° – 40° = 30°

Also, ∠AOC + ∠BOE + ∠COE = 180° ( AOB is a  straight line)

⇒70° + ∠COE = 180° [from (1)]

⇒ ∠COE = 180° – 70° = 110°

Now, reflex ∠COE = 360° – 110° = 250°

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

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

In the figure, lines XY and MN intersect at O and ∠ POY = 90°.

Also, given a : b = 2 : 3

Let a = 2x and b = 3x.

Since, ∠XOM + ∠POM + ∠POY = 180°(Linear pair axiom)

⇒ 3x + 2x + 90° = 180°

⇒ 5x = 180° – 90°

⇒ x = 90°/5 = 18°

∴ ∠XOM = b = 3x = 3 × 18° = 54°

and ∠POM = a = 2x = 2 × 18° = 36°

Now, ∠XON = c = ∠MOY = ∠POM + ∠POY (Vertically opposite angles)

= 36° + 90° = 126°

Hence, c = 126°

1. In Fig., ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT. ∠PQS + ∠PQR = 180° …………….(1)(Linear pair axiom)

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

But, ∠PQR = ∠PRQ (Given)

∴ From (1) and (2)

∠PQS = ∠PRT

1. In Fig., if x + y = w + z, then prove that AOB is a line. Sum of all angles in a circle always 360°

Hence, ∠AOC + ∠BOC + ∠DOB + ∠AOD = 360°

⇒ x + y + w + z = 360°

⇒ x + y + x + y = 360°

Given that x = y = w + z

⇒ 2w + 2z = 360°

⇒ 2(w + z) = 360°

⇒ w + z = 180° (linear pair)

ie., ∠BOD + ∠AOD = 180°

If the sum of two adjacent angles is 180° then the non common arms of the angles form a line.

Therefore, AOB is a line.

1. In Fig., POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = 1/2 (∠ QOS – ∠ POS). Solution:

Since, OR⊥PQ
⇒ ∠QOS = 90°
Since, PQ is a line OR stands on it
Therefore, ∠POR+∠QOR = 180 °      (linear pair)
∠POR = 90 °
Since, ∠POS + ∠SOR = 90 °………..(i)
∠QOR = 90 °
∠QOS – ∠ROS = 90 °   ……………….. (ii)

From (i) and (ii)

∠POS + ∠SOR = ∠QOS -∠POS
∠ROS + ∠ROS = ∠QOS – ∠POS
2∠ROS = ∠QOS-∠POS

∠ROS = 1/2 (∠QOS – ∠POS)
Hence proved

1. 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: Given, ∠XYZ = 64° and QY bisects ∠ZYP

⇒∠XYZ + ∠ZYP = 180° [LINEAR PAIR]

⇒64° + ∠ZYP = 180°

⇒∠ZYP = 180° – 64°

⇒∠PYQ + ∠ZYQ = 116°

⇒2∠PYQ = 116°

⇒∠PYQ = 58°

Again, ∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58°

= 122°