Question

In 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

REF. Image

$\angle SOR=\angle TOR$

$\angle ROS=\angle ROT$

$\angle SOT=\angle QOS-\angle QOT$

$\angle SOT=\angle QOS-\angle POS$. ................. $(\angle POS=\angle QOT={90}^{\circ }-\theta )$

$2\angle ROS=\angle QOS-\angle POS$ ......................... $(\angle SOT=\angle ROS+\angle ROT)$

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