Question

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

Answer 1

Given
$\angle ABC={69}^{\circ },\angle ACB={31}^{\circ }$
In the triangle ABC,
$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$ [Angle sum property of Triangle]
${69}^{\circ }+{31}^{\circ }+\angle BAC={180}^{\circ }$
${100}^{\circ }+\angle BAC={180}^{\circ }$
$\therefore \angle BAC={180}^{\circ }-{100}^{\circ }$
$\Rightarrow \angle BAC={80}^{\circ }$
$\angle BAC=\angle BDC={80}^{\circ }$ [Angles in the same segment are equal]
$\therefore \angle BDC={80}^{\circ }$