Question

In the adjacent figure ray OS stands on a line PQ. Ray OR and ray OT are angle bisectors of $\angle POS$ and $\angle SOQ$ respectively. Find $\angle ROT$.

Answer 1

Ray OS stands on the line PQ.

Therefore, $\angle POS+\angle SOQ={180}^0$ (Linear pair)

Let $\angle SOQ=x^0$

Thus $\angle POS={180}^0-\angle SOQ$

$\angle POS={180}^0-x^0$

Now, ass ray $OR$ and ray $OT$ are angle bisectors so

$\angle SOT=\frac{1}{2}\times \angle SOQ$

Thus, $\angle SOT=\frac{1}{2}\times x^0$

And $\angle ROS=\frac{1}{2}\times \angle POS$

Thus, $\angle ROS=\frac{1}{2}\times ({180}^0-x^0)$

$\angle ROS={90}^0-\frac{x^0}{2}$

Here $\angle ROT=\angle ROS+\angle SOT$

$\angle ROT={90}^0-\frac{x^0}{2}+\frac{x^0}{2}$

$\angle ROT={90}^0$.