Question

In $\Delta ABC$, AD bisects $\angle BAC$ and $AD=DC.$ If $\angle ADB={100}^0,$ then find $\angle ABD.$

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

The correct option is A ${30}^0$

As $AD=DC$

So, $\angle C=\angle DAC$ (Angle corresponding to equal sides)

and, $\angle C+\angle DAC=100°$ (Exterior angle)

So, $\angle C=\angle DAC=50°$

Now, $\angle A=2\times \angle DAC=100°$

and in $△ABC$,

$\angle A+\angle B+\angle C=180°$

So, $\angle B=180°-100°-50°=30°$