Question

In the given figure, ABCD is a cyclic quadrilateral in which $\angle BAD={120}^{\circ }$. Find $\angle BCD$.

• A. ${240}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

Sum of the opposite angles in a cyclic quadrilateral $={180}^{\circ }$
$\therefore \angle BAD+\angle BCD={180}^{\circ }$
$\angle BCD={180}^{\circ }-\angle BAD$
$={180}^{\circ }-{120}^{\circ }={60}^{\circ }$