Question

In the fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and QR. Prove that

Answer 1

POQ is a straight line. [Given]

∴ ∠POS + ∠ROS + ∠ROQ = 180°

But OR ⊥ PQ

∴ ∠ROQ = 90°

⇒ ∠POS + ∠ROS + 90° = 180°

⇒ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS … (1)

Now, we have ∠ROS + ∠ROQ = ∠QOS

⇒ ∠ROS + 90° = ∠QOS

⇒ ∠ROS = ∠QOS – 90° ……(2)

Adding (1) and (2), we have

2 ∠ROS = (∠QOS – ∠POS)

∴ ∠ROS = 

Answer 2

GIVEN :-

• POQ is a line. Ray OR is perpendicular to line PQ.

• OS is another ray lying between rays OP and QR.

TO PROVE :-

• $\rm{\angle{ROS} = \dfrac{1}{2}(\angle{QOS}- \angle{POS}}$

PROOF :-

Given that OR is Perpendicular to PQ

$\implies\rm{\angle{POR} = \angle{ROQ} =90^{\circ}}$

$\implies\rm{\angle{POS} +\angle{ROS} = 90^{\circ}}$

$\implies\rm{\angle{ROS} = 90^{\circ} -\angle{POS}}$

Adding ROS to both sides, we have

$\implies\rm{\angle{ROS} +\angle{ROS} = (90^{\circ}+\angle{ROS})-\angle{POS}}$

$\implies\rm{2\angle{ROS} = \angle{QOS} -\angle{POS}}$

So,

$\implies\rm{\angle{ROS}=\dfrac{1}{2}(\angle{QOS} - \angle{POS})}$.