Question

In $△$ ABC, $\angle A$ = ${100}^{\circ }$ , AD bisects $\angle A$ and AD $\perp$ BC. Find $\angle$ B

Answer 1

In $△ABC,\angle BAD+\angle ADB+\angle DBA={180}^{\circ }$ by angle sum property of a triangle.

${50}^{\circ }+{90}^{\circ }+\angle DBA={180}^{\circ }$

$\Rightarrow \angle DBA={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

$\therefore \angle B={40}^{\circ }$