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. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${75}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is A

${90}^{\circ }$

Here, $\angle POS$ and $\angle SOQ$ are adjacent angles with non common arms OP and OQ forming a line.

$\Rightarrow \angle POS$ + $\angle SOQ$ = ${180}^{\circ }$ (Linear pair)

$\angle POS$ = $x$ (Given)

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

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

Now, ray OR bisects $\angle POS$.

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

$\Rightarrow$ $\angle ROS=\frac{x}{2}$

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

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

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

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

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

$\Rightarrow \angle ROT$= ${90}^{\circ }$