Question

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

Answer 1

Given : PQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

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

Proof:

Since OR is perpendicular to $PQ$,

$\angle ROP=\angle ROQ......\left(i\right)$

Also ,we can see that $\angle ROP=\angle ROS+\angle SOP$ and $\angle ROQ=\angle QOS-\angle ROS$

So, putting these values in equation (i),we get

$\angle ROS+\angle SOP=\angle QOS-\angle ROS$

Adding $\angle ROS$ on both sides,

$\angle ROS+\angle SOP+\angle ROS=\angle QOS-\angle ROS+\angle ROS$

$2\angle ROS+\angle SOP=\angle QOS$

Subtracting $\angle SOP$ from both the sides,

$2\angle ROS+\angle SOP-\angle SOP=\angle QOS-\angle SOP$

$2\angle ROS=\angle QOS-\angle SOP=$

Dividing both sided by $2$,

$\angle ROS=(\angle QOS-\angle SOP)/2$