Question

In the given figure, $OP$ bisect $\angle BOC$ and $OQ$ bisect $\angle AOC$. Then $\angle POQ$ is equal to

• A. ${90}^{\circ }$
• B. ${120}^{\circ }$
• C. ${60}^{\circ }$
• D. ${100}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$

R.E.F image

$\angle BOP=\frac{1}{2}\angle BOC$ $\left\{Given\right\}$

$\angle AOQ=\frac{1}{2}\angle AOC$ $\left\{Given\right\}$

$\angle BOC+\angle AOC={180}^0$ $\left\{condition\right\}$

$2\angle BOP+2\angle AOQ={180}^0$ Since

$\angle BOP+\angle AOQ={90}^0$ $\angle BOP=\angle POC$

$\angle POC+\angle COQ={90}^0$ $\angle QOC=\angle AOQ$

$\angle POQ={90}^0$