Question

In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P. If ∠C = 100° and ∠D = 60°, find the measure of ∠APB.

Answer 1

Sum of the angles of a quadrilateral is 360°.
$$\begin{gathered} \therefore \angle A+\angle B+{60}^o+{100}^o=360° \\ \angle A+\angle B=360-100-60=200° \\ or \\ \frac{1}{2}\left(\angle A+\angle B\right)=100°...\left(1\right) \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{In}△APB: \\ \frac{1}{2}\left(\angle A+\angle B\right)+\angle P=180°(\mathrm{because}AP\mathrm{and}PB\mathrm{are}\mathrm{bisectors}\mathrm{of}\angle A\mathrm{and}\angle B) \\ U\mathrm{sing}\mathrm{equation}\left(1\right): \\ 100°+\angle P=180° \\ \Rightarrow \angle P=80° \end{gathered}$$

∴ $\angle APB=80°$