Question

$POQ$ is a line ray $OR$ is a perpendicular to line $PQ$. $OS$ is another ray lying between ray $OP$ and $OR$. Prove that $\angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$

Answer 1

Given:- $OR\perp PQ$

To prove:-

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

Proof:-

$\because OR\perp PQ$

$\therefore \angle ROP=90°$

$\angle ROQ=90°$

$\therefore$ We can say that,

$\angle ROP=\angle ROQ$

$\angle POS+\angle ROS=\angle ROQ[\because \angle ROP=\angle POS+\angle ROS]$

$\Rightarrow \angle POS+\angle ROS=\angle QOS-\angle ROS[\because \angle ROQ=\angle QOS-\angle ROS]$

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

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

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