Question

Question 5
In the given 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

It is given that $OR\perp PQ$.
$\therefore \angle POR={90}^{\circ }$
$\angle POS+\angle SOR={90}^{\circ }$
$\angle ROS={90}^{\circ }-\angle POS$ ………..(1)
$\angle QOR={90}^{\circ }(AsOR\perp PQ)$
$\angle ROS=\angle QOS-\angle QOR$
$\angle ROS=\angle QOS-{90}^{\circ }$………..(2)
On adding equation (1) and (2), we obtain,
$\angle ROS+\angle ROS={90}^{\circ }-\angle POS+\angle QOS–{90}^{\circ }$
$2\angle ROS=\angle QOS-\angle POS$
$\angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$