Question

In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
$\angle ROS = \dfrac{1}{2}\left( {\angle QOS - \angle POS} \right)$

Answer 1

Hint: We will first prove that $\angle ROP$is 90° and then we can expand $\angle ROP$in terms of the angles ROS and POS. and upon simplifying the obtained equation, we will prove that $\angle ROS = \dfrac{1}{2}\left( {\angle QOS - \angle POS} \right)$

Complete step by step answer:

We are given that POQ is a line and a ray OR is perpendicular to it. OS is some other ray which is also located between rays OP and OR.
We need to prove that $\angle ROS = \dfrac{1}{2}\left( {\angle QOS - \angle POS} \right)$
Let us first see the figure:

Here, as we can see that OR is perpendicular to PQ. So, if $\angle $ROQ = 90°, then$\angle $ROP must be 90 °.
Therefore, we can say that $\angle $ROP = $\angle $ROQ
Now, from the figure, we can see that $\angle $ROP = $\angle $ROS +$\angle $POS
$ \Rightarrow $$\angle $POS + $\angle $ROS = $\angle $ROQ
Also, $\angle $ROQ = $\angle $QOS - $\angle $ROS
$ \Rightarrow $$\angle $POS + $\angle $ROS = $\angle $QOS - $\angle $ROS
Rearranging the terms, we get
$ \Rightarrow $$\angle $ROS + $\angle $ROS = $\angle $QOS -$\angle $POS
$ \Rightarrow $2($\angle $ROS) = $\angle $QOS -$\angle $POS
$ \Rightarrow $$\angle $ROS = $\dfrac{1}{2}$($\angle $QOS -$\angle $POS)
Hence, we have proved the required relation that $\angle $ROS = $\dfrac{1}{2}$($\angle $QOS -$\angle $POS)

Note: In such questions, you may get confused while proving the relation between two angles like in this question, we deduced a relation between $\angle $ROP and $\angle $ROQ as OR was a perpendicular ray on PQ. You may go wrong while simplifying the $\angle $ROP in terms of $\angle $POS and $\angle $ROS in this question.