Question

In ∆ABC, ∠A = 100°, AD bisects ∠A and AD ⊥ BC. Find ∠B.

Answer 1

$$\begin{gathered} \mathrm{Consider}△\mathrm{ABD}. \\ \angle \mathrm{BAD}=\frac{100°}{2}(\because \mathrm{AD}\mathrm{bisects}\angle \mathrm{A}) \\ \Rightarrow \angle \mathrm{BAD}=50° \\ \angle \mathrm{ADB}=90°(\because \mathrm{AD}\perp \mathrm{BC}) \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Thus}, \\ \angle \mathrm{ABD}+\angle \mathrm{BAD}+\angle \mathrm{ADB}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{ABD}) \\ \mathrm{Or}, \\ \angle \mathrm{ABD}+50°+90°=180° \\ \angle \mathrm{ABD}=180°-140° \\ \angle \mathrm{ABD}=40° \end{gathered}$$