In Fig. 16.21, the bisectors of ∠A and ∠B meet at a point P.

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ}$

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+100^{\circ}+50^{\circ}=360^{\circ}$

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}=210^{\circ} \quad \ldots(\mathrm{i})$

In $\triangle \mathrm{APB}$, we have $:$

$\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{B}+\angle \mathrm{APB}=180^{\circ}$

$\Rightarrow \angle \mathrm{APB}=180^{\circ}-\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B})$

From (i), we get:

$\Rightarrow \angle \mathrm{APB}=180^{\circ}-\left(\frac{1}{2} \times 210^{\circ}\right)$

$\therefore \angle \mathrm{APB}=75^{\circ}$