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

(N/A) $POQ$ is a straight line. [Given]
$\therefore \angle POS + \angle ROS + \angle ROQ = 180^o$
But $OR \perp PQ$,therefore $\angle ROQ = 90^o$.
Substituting $\angle ROQ = 90^o$ in the equation:
$\angle POS + \angle ROS + 90^o = 180^o$
$\Rightarrow \angle POS + \angle ROS = 90^o$ --- $(1)$
Now,we have $\angle QOS = \angle ROQ + \angle ROS$
Since $\angle ROQ = 90^o$,we have:
$\angle QOS = 90^o + \angle ROS$
$\Rightarrow 90^o = \angle QOS - \angle ROS$ --- $(2)$
From $(1)$ and $(2)$,we have:
$\angle POS + \angle ROS = \angle QOS - \angle ROS$
$\Rightarrow \angle ROS + \angle ROS = \angle QOS - \angle POS$
$\Rightarrow 2 \angle ROS = \angle QOS - \angle POS$
$\therefore \angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$