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In this question, we are supposed to prove that $ \angle ROS=\dfrac{1}{2}\left( \angle QOS-\angle POS \right) $ where POQ is a line and Ray OR is perpendicular to the line P and also OS is another ray lying between rays OP and OR.

Now, for this type of question, we will use various angles from the figure to prove the equation mentioned in the question.

So, we can clearly see from the figure that:

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

So, by rearranging the above equation, we get:

$ \angle QOS-\angle ROS={{90}^{\circ }}....\left( i \right) $

Then, we also know that the angle formed on the straight line is always supplementary in nature, that is it adds equal to $ {{180}^{\circ }} $ .

So, by using this condition, we get:

$ \angle QOS+\angle POS={{180}^{\circ }} $

The, we can write 180 as multiplication of 2 and 90 as:

$ \angle QOS+\angle POS=2\times {{90}^{\circ }} $

The, by substituting the value of $ {{90}^{\circ }} $ from equation (i) in the above mentioned expression, we get:

$ \angle QOS+\angle POS=2\times \left( \angle QOS-\angle ROS \right) $

Now, by solving the above expression, we get:

$ \begin{align}

& \angle QOS+\angle POS=2\angle QOS-2\left( \angle ROS \right) \\

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

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

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

\end{align} $

So, we get the condition $ \angle ROS=\dfrac{1}{2}\left( \angle QOS-\angle POS \right) $ as proved.

Hence, the condition is proved.