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 \(∠\)ROS = \(\frac{1}{2}\) (\(∠\)QOS – \(∠\)POS)

Answer 1

We can rewrite the following as: 

\[ \angle ROP = \angle ROQ = 90^\circ \]

Since \( \angle ROP = \angle ROQ \), we can write:

\[ \angle POS + \angle ROS = \angle ROQ \]

Now, substituting for \( \angle ROQ \), we get:

\[ \angle POS + \angle ROS = \angle QOS - \angle ROS \]

Rearranging the terms:

\[ \angle SOR + \angle ROS = \angle QOS - \angle POS \]

Thus, we obtain:

\[ 2 \times \angle ROS = \angle QOS - \angle POS \]

Therefore:

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