Question

From the adjoining figure, find $\angle$ BDC and $\angle$ ABD.

Answer 1

$\angle B=\angle C={65}^{\circ }$
$\angle B+\angle C+\angle A={180}^{\circ }$ [ Angle property]
$Also\angle B=\angle C$ [ Isoceles triangle]
Thus $\angle B+\angle C+\angle A=180$
$\Rightarrow \angle B+\angle B+50=180$
$\Rightarrow 2\angle B=130$
$\Rightarrow \angle B=\frac{130}{2}={65}^0=\angle C$
In $\Delta BCD$
$\angle B=\angle D$ [Isoceles triangle]
Thus
$\angle C+\angle B+\angle D={180}^{\circ }$ [Angle property]
$\angle C+\angle B+\angle D=180$
$2\angle B=180-\angle C$
$\Rightarrow \angle B=180-90=\frac{90}{2}$
$\Rightarrow \angle B={45}^{\circ }$
Thus $\angle BDC={45}^{\circ }$
$\angle ABD=65+45={110}^{\circ }$.