Question

In figure, ray $OS$ stands on a line $POQ$. Ray $OR$ and ray $OT$ are angle bisectors of $\angle POS$ and $\angle SOQ$, respectively. If $\angle POS=x,$ find $\angle ROT$.

Answer 1

Given: Ray $OS$ stands on a line $POQ$,

Since, $POQ$ is a straight line

$\Rightarrow \angle POS+\angle SOQ={180}^{\circ }$

$\Rightarrow x+\angle SOQ={180}^{\circ }$

$\Rightarrow \angle SOQ={180}^{\circ }-x$

Now, ray $OR$ bisects $\angle POS$

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

Similarly, $\angle SOT=\frac{1}{2}\times \angle SOQ=\frac{1}{2}\times (180-x^{\circ })$

$={90}^{\circ }-\frac{x}{2}$

$\therefore \angle ROT=\angle ROS+\angle SOT=\frac{x}{2}+{90}^{\circ }-\frac{x}{2}={90}^{\circ }$

Hence, $\angle ROT={90}^{\circ }$.