Question

please solve this question​

Answer 1

Given :

• ∠ROQ = 90°

To Prove :

• ∠ROS = $\dfrac{1}{2}$(∠QOS-∠POS)

Solution :

As POQ is a line . So ,

$\longmapsto\tt{\angle{POS}+\angle{ROS}+\angle{ROQ}=180\degree}$

$\longmapsto\tt{\angle{POS}+\angle{ROS}+90\degree=180\degree}$

$\longmapsto\tt{\angle{POS}+\angle{ROS}=180\degree-90\degree}$

$\longmapsto\tt\bf{\angle{POS}+\angle{ROS}=90\degree\:---(1)}$

Also ,

$\longmapsto\tt{\angle{QOS}=\angle{ROS}+\angle{ROQ}}$

$\longmapsto\tt{\angle{QOS}=\angle{ROS}+90\degree}$

$\longmapsto\tt\bf{\angle{QOS}-\angle{ROS}=90\degree\:---(2)}$

By Equation 1 nd 2 :

$\longmapsto\tt{\angle{POS}+\angle{ROS}=\angle{QOS}-\angle{ROS}}$

$\longmapsto\tt{\angle{ROS}+\angle{ROS}=\angle{QOS}-\angle{POS}}$

$\longmapsto\tt\bf{\angle{ROS}=\dfrac{1}{2}(\angle{QOS}-\angle{POS})}$

HENCE PROVED

Answer 2

Given:-

•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)

Hence verified!