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

Given $OR\perp PQ$.

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

Proof,

$\angle ROS=\angle ROP-\angle POS⟶\left(i\right)$

also, $\angle ROS=\angle QOS-\angle ROQ⟶\left(ii\right)$

Now, $\angle ROQ=\angle ROP$ [Both ${90}^0$]

$\therefore$ (ii) $\Rightarrow$ $\angle ROS=\angle QOS-\angle ROP⟶\left(iii\right)$

$\left(i\right)+\left(iii\right)$ $\Rightarrow$ $2\angle ROS=\angle QOS-\angle POS$

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

Hence Proved.