In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P.

Sum of the angles of a quadrilateral is 360°.

$\therefore \angle A+\angle B+60^{\circ}+100^{\circ}=360^{\circ}$

$\angle A+\angle B=360-100-60=200^{\circ}$

or

$\frac{1}{2}(\angle A+\angle B)=100^{\circ} \quad \ldots(1)$

Sum of the angles of a triangle is $180^{\circ}$.

In $\triangle A P B$ :

$\frac{1}{2}(\angle A+\angle B)+\angle P=180^{\circ} \quad$ (because $A P$ and $P B$ are bisectors of $\angle A$ and $\angle B$ )

$U$ sing equation (1):

$100^{\circ}+\angle P=180^{\circ}$

$\Rightarrow \angle P=80^{\circ}$

$\therefore \angle A P B=80^{\circ}$