Question

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

Answer 1

Ray OR $\perp ROQ$. OS is another ray lying between OP and OR.

To prove: $\angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$
Proof: $\because RO\perp POQ$
$\therefore \angle POR={90}^{\circ }$
$\Rightarrow \angle POS+\angle ROS={90}^{\circ }$ ...(i)
$\Rightarrow \angle ROS={90}^{\circ }-\angle POS$
But $\angle POS+\angle QOS={180}^{\circ }$ (Linear pair)
=$2(\angle POS+2\angle ROS)$ [From (i)]
$\angle POS+\angle QOS=2\angle ROS+2\angle POS$
$\Rightarrow 2\angle ROS=\angle POS+\angle QOS-2\angle POS$
=$\angle QOS-\angle POS$
$\therefore ROS=\frac{1}{2}(\angle QOS-\angle POS)$