Question

In the figure, $\angle ABC={69}^o,\angle ACB={31}^o$, find $\angle BDC$.

Answer 1

$\angle BAC=\angle BDC$ ...Angles in the segment of the circle

In $\Delta ABC$,

$\Rightarrow$$\angle BAC+\angle ABC+\angle ACB=180°$ ...Sum of the angles in a triangle

$\Rightarrow \angle BAC+69°+31°=180°$

$\Rightarrow \angle BAC=180°-100°$

$\Rightarrow \angle BAC=80°$

Thus, $\angle BDC=80°$