Question

In fig. 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

$\angle ROS={90}^{\circ }-\angle POS-\left(i\right)$

$\angle QOS=\angle QOR+\angle ROS={90}^{\circ }+\angle ROS$

$\Rightarrow {90}^{\circ }=\angle QOS-\angle ROS-\left(ii\right)$

Substituting $\left(ii\right)$ in $\left(i\right)$ we get

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

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

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

Hence proved.