Question

In figure $POQ$ is a line. Ray $OR$ is perpendicular to the line $PQ,OS$ is another ray lying between rays $OP$ and $QR$. Prove that $\angle ROS=\frac{1}{2}(\angle QOS-\angle PQS)$

Answer 1

To prove : $\angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$

Prrof :

$\angle ROP={90}^0$ ( $\because$ $\angle ROQ={90}^0;\therefore \angle ROP={180}^0-\angle ROQ$)

$\Rightarrow \angle ROS+\angle POS={90}^0$

$\Rightarrow \angle ROS+\angle POS=\angle ROQ$ ( $\because$ $\angle ROS={90}^0$)

$\Rightarrow \angle ROS+\angle POS=\angle QOS-\angle ROS$

$\Rightarrow 2\angle ROS=\angle QOS-\angle POS$

$\Rightarrow \angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$

Hence proved.