Question

In the figure, $\angle ABD=3\angle DAB$ and $\angle BDC={96}^o$, Find $\angle ABD$

Answer 1

In $△ABC$

$\angle ABD=3\angle DAB$ (given)

and $\angle BDC={96}^{\circ }$

$\angle BDC+\angle BDA={180}^{\circ }$ (linear pair)

$\Rightarrow {96}^{\circ }+\angle BDA={180}^{\circ }$

$\Rightarrow \angle BDA={180}^{\circ }-{96}^{\circ }={84}^{\circ }$

In $△ABD,\angle ABD+\angle DAB+\angle BDA={180}^{\circ }$ (angle sum property)

$\Rightarrow 3\angle DAB+\angle DAB+{84}^{\circ }={180}^{\circ }$

$\Rightarrow 4\angle DAB={180}^{\circ }-{84}^{\circ }={96}^{\circ }$

$\Rightarrow \angle DAB=\frac{{96}^{\circ }}{4}={24}^{\circ }$

and $\angle ABD=3\angle DAB=3\times {24}^{\circ }={72}^{\circ }$