Question

In the figure below, 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 $\angle ROT$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Ray OS stands on the line POQ, so that,

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

But, $\angle POS=x$

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

$\angle SOQ={180}^{\circ }–x$

Now, ray OR bisects POS.

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

Similarly, $SOT=\frac{1}{2}\times \angle SOQ\Rightarrow \frac{1}{2}\times ({180}^{\circ }-x)=({90}^{\circ }–\frac{x}{2})$

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