Question

In the adjacent figure, the bisectors of $\angle A$ and $\angle B$ meet at a point P. If $\angle C={100}^0$ and $\angle D={60}^0$, find the measure of $\angle APB$.

• A. ${100}^{\circ }$
• B. ${80}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B ${80}^{\circ }$


In quadrilateral ABCD
$\angle C={100}^0,\angle D={60}^0$
$\angle A+\angle B+\angle C+\angle D={360}^0$
(Angle sum property of a quadrilateral)
$\therefore \angle A+\angle B={360}^0-({100}^0+{60}^0)$
= ${360}^0-{160}^0={200}^0$

But AP and BP are the bisectors of $\angle A$ and $\angle B$.
$\therefore \frac{1}{2}(\angle A+\angle B)=\frac{1}{2}\times {200}^0={100}^0$
i.e. $\angle PAB+\angle PBA={100}^0$

But in $\Delta APB$,
$\angle PAB+\angle PBA+\angle APB={180}^0$
$\Rightarrow {100}^0+\angle APB={180}^0$
$\angle APB={180}^0-{100}^0={80}^0$
Thus, $\angle APB={80}^0$