Question

In the given 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

From the figure

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

$\angle QOS+\angle POS={180}^{\circ }=2\left({90}^{\circ }\right)$

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

$⟹2\angle ROS=\angle QOS-\angle POS$

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