Question

$ABCD$ is a cyclic quadrilateral. If $\angle BAC={50}^{\circ }$ and $\angle DBC={60}^{\circ }$ then find $\angle BCD$

Answer 1

Here $\angle BDC=\angle BAC={50}^{\circ }$ since angles in same segment are equal

In $△BCD$, we have

$\angle BCD={180}^{\circ }-(\angle BDC+\angle DBC)$

$={180}^{\circ }-({50}^{\circ }+{60}^{\circ })$

$={180}^{\circ }-{110}^{\circ }={70}^{\circ }$

$\therefore \angle BCD={70}^{\circ }$