Question

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

Answer 1

In $△ABC$

$\angle BAC+\angle ABC+\angle ACB={180}^o$ (Angle sum property)

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

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

For segment $BADCB$,

$VBDC$ and $\angle BAC$ are angles in the same segment. So, they must be equal

$\therefore$ $\angle BDC=\angle BAC$ (Angles in the same segment are equal)

$\angle BDC={80}^o$