Question

In the given figure, $\angle ABC={69}^{\circ },\angle ACB={31}^{\circ }$.Find $\angle BDC$

Answer 1

In $△ABC,$

$\angle BAC+\angle ABC+\angle ACB={180}^{\circ }$

$\Rightarrow \angle BAC+{69}^{\circ }+{31}^{\circ }={180}^{\circ }$

$\Rightarrow \angle BAC={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

For segment $BADCB,$

$\angle BDC$ and $\angle BAC$ are angles in the same segment.

So, they must be equal.

$\therefore \angle BDC=\angle BAC$

$\Rightarrow \angle BDC={80}^{\circ }$