Question

In the following figure, $\angle ABD=3\angle DAB$ and $\angle ADC={120}^{\circ }$ then find $\angle ABD$.

Answer 1

Given $\angle ADC={120}^{\circ }$

AS $\angle BDC={180}^{\circ }$ and

$\angle BDC=\angle BDA+\angle ADC$

$\Rightarrow {180}^{\circ }=\angle BDA+{120}^{\circ }$

$\Rightarrow \angle BDA={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

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

Given:$\angle ABD=3\angle DAB$

$\Rightarrow {60}^{\circ }+3\angle DAB+\angle BAD={180}^{\circ }$

$\Rightarrow 4\angle BAD={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

$\therefore \angle BAD=\frac{{120}^{\circ }}{4}={30}^{\circ }$

From above $\angle ABD=3\angle BAD=3\times {30}^{\circ }={90}^{\circ }$