Question

In figure above, the bisectors of $\angle P$ and $\angle Q$ meet at a point O. If $\angle R={100}^{\circ }$ and $\angle S={50}^{\circ }$, find $\angle POQ$.

Answer 1

We know that sum of interior angles of quadrilateral is ${360}^o$

$\angle P+\angle Q+\angle R+\angle S={360}^o$

$\angle P+\angle Q+100+50=360$

$\angle P+\angle Q={210}^o$ ……………..$\left(1\right)$

OP and OQ are bisectors of angles P and Q,

$\angle P=2\angle OPQ$ and $\angle Q=2\angle OQP$

So from (i), we have

$2\angle OPQ+2\angle OQP={210}^o$

$2(\angle OPQ+\angle OQP)={210}^o$

$\angle OPQ+\angle OQP={105}^o$ …………….$\left(2\right)$

Consider triangle POQ,

$\angle POQ+\angle OPQ+\angle OQP={180}^o$

$\angle POQ+{105}^o={180}^o$ [using equation $\left(3\right)$]

$\angle POQ={180}^o-{105}^o$

$\angle POQ={75}^o$.