Question

In Fig.$ POQ$ is a line. Ray $ OR$ is perpendicular to line $ PQ. OS $is another ray lying between rays $ OP$and $ OR.$Prove that $ ROS=\frac{1}{2}(QOS-POS).$

Answer 1

Solve for value of the required angle

Given,

• $POQ$is a line.
• Ray $OR$is perpendicular to line $PQ.$
• $OS$ is another ray lying between rays $OP$and $OR.$

$POQ$is a straight line. The sum of all angles made on it is $180°.$

$\Rightarrow \angle POS+\angle ROS+\angle ROQ=180°$

$\Rightarrow$ $\angle POS+\angle ROS+90=180$ [given $\angle ROQ=90°$]

$\Rightarrow$ $\angle POS+\angle ROS=90$

$\Rightarrow$ $\angle ROS=90-\angle POS...\left(i\right)$

$\Rightarrow$ $\angle ROS+\angle ROQ=\angle QOS$ [from figure]

$\Rightarrow$ $\angle ROS=\angle QOS-90°...\left(ii\right)$

On adding both the equations $\left(i\right)+\left(ii\right)$

$\Rightarrow$ $\angle ROS+\angle ROS=90°-\angle POS+\angle QOS-90°$

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

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

Hence proved that $ROS=\frac{1}{2}(QOS-POS).$