Question

Question 40

In $\Delta ABC$, $\angle A={100}^{\circ }$, AD bisects $\angle A$ and $AD\perp BC$. Then, $\angle B$ is equal to

a) ${80}^{\circ }$
b) ${20}^{\circ }$
c) ${40}^{\circ }$
d) ${30}^{\circ }$

Answer 1

Given, $\angle BAD=\angle DAC={50}^{\circ }$ [$\because$ AD bisects $\angle A$ and $\angle A={100}^{\circ }$]
and $\angle BDA=\angle ADC={90}^{\circ }$ $[\because AD\perp BC]$


Now, in $\Delta ABD$,
$\angle ABD+\angle BAD+\angle BDA={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow \angle ABD+{50}^{\circ }+{90}^{\circ }={180}^{\circ }\Rightarrow \angle ABD+{140}^{\circ }={180}^{\circ }\Rightarrow \angle ABD={180}^{\circ }-{140}^{\circ }\Rightarrow \angle ABD={40}^{\circ }$