Question

5. 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
S
ZROS =
(Z QOS-ZPOS).
2​

Answer 1

Given :-

• POQ is a line
• ray OR is perpendicular to PQ
• OS is another ray lying b/w OP and OR

To prove :-

• 2 ∠ ROS = (∠ QOS - ∠ POS)

Proof :-

Taking RHS

RHS = ∠ QOS - ∠ POS

( ∵ ∠ QOS = ∠ ROQ + ∠ ROS )

RHS = ∠ ROQ + ∠ ROS - ∠ POS

( ∵ OR ⏊ PQ, ∴ ∠ROQ = ∠POR = 90°)

RHS = ∠ POR + ∠ ROS - ∠ POS

RHS= ( ∠ POR - ∠ POS ) + ∠ ROS

RHS = ∠ ROS + ∠ ROS

RHS = 2 ∠ ROS = LHS

PROVED.

Answer 2

Answer:

Step-by-step explanation:

It is given that OR is perpendicular to  PQ

So that ∠POR = 90°

sum of angle in linear pair always equal to 180°

∠POS + ∠SOR + ∠POR = 180°

Plug ∠POR = 90°

90°+∠SOR + ∠POR = 180°

∠SOR + ∠POR = 90°

∠ROS = 90° − ∠POS                       … (1)

∠QOR = 90°

Given that OS is another ray lying between rays

OP and OR  so that  

∠QOS − ∠ROS = 90°

∠ROS = ∠QOS − 90°                              … (2)

On adding equations (1) and (2), we obtain

2 ∠ROS = ∠QOS − ∠POS

∠ROS = 1/2(∠QOS − ∠POS)