∠POY = 90° and a : b = 2 : 3

A/q,

∠POY + a + b = 180°

⇒ 90° + a + b = 180°

⇒ a + b = 90°

Let a be 2x then will be 3x

2x + 3x = 90°

⇒ 5x = 90°

⇒ x = 18°

∴ a = 2×18° = 36°

and b = 3×18° = 54°

also,

b + c = 180° (Linear Pair)

⇒ 54° + c = 180°

⇒ c = 126°

∠PQR = ∠PRQ

To prove,

∠PQS = ∠PRT

A/q,

∠PQR +∠PQS = 180° (Linear Pair)

⇒ ∠PQS = 180° - ∠PQR --- (i)

also,

∠PRQ +∠PRT = 180° (Linear Pair)

⇒ ∠PRT = 180° - ∠PRQ

⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)

From (i) and (ii)

∠PQS = ∠PRT = 180° - ∠PQR

Therefore, ∠PQS = ∠PRT

x + y = w + z

To Prove,

AOB is a line or x + y = 180° (linear pair.)

A/q,

x + y + w + z = 360° (Angles around a point.)

⇒ (x + y) + (w + z) = 360°

⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)

⇒ 2(x + y) = 360°

⇒ (x + y) = 180°

Hence, x + y makes a linear pair. Therefore, AOB is a staright line.

In Fig. 6.17, 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).

Given,

OR is perpendicular to line PQ

To prove,

∠ROS = 1/2(∠QOS – ∠POS)

A/q,

∠POR = ∠ROQ = 90° (Perpendicular)

∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS --- (i)

∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)

Subtracting (ii) from (i)

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

⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS

⇒ ∠QOS - ∠POS = 2∠ROS

⇒ ∠ROS = 1/2(∠QOS – ∠POS)

Hence, Proved.