Question

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

Answer 1

Given,
$OR$ is perpendicular to line $PQ$

To prove: $\angle ROS=\frac{1}{2}(\angle QOS–\angle POS)$

According to the question,

$\angle POR=\angle ROQ={90}^{\circ }$ $\mid$ Perpendicular

$\angle QOS=\angle ROQ+\angle ROS={90}^{\circ }+\angle ROS$ --- (i)

$\angle POS=\angle POR-\angle ROS={90}^{\circ }-\angle ROS$--- (ii)

Subtracting (ii) from (i)

$\angle QOS-\angle POS={90}^{\circ }+\angle ROS-({90}^{\circ }-\angle ROS)$

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

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

$\Rightarrow \angle ROS=\frac{1}{2}(\angle QOS–\angle POS)$ $\left[henceproved\right]$