Question

In the given figure, $ABCD$ is a cyclic quadrilateral in which $\angle BAD={120}^o$. Then $m\angle BCD$ is:

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

Answer 1

Answer: (B)

${60}^o$

Given, $ABCD$ is a cyclic quadrilateral

and $\angle BAD={120}^o$.

Since, $ABCD$ is a cyclic quadrilateral,

$\therefore \angle BCD+\angle BAD={180}^o$ ...[Opposite angles of cyclic quadrilateral are supplementary].

$⟹\angle BCD={180}^o-\angle BAD={180}^o-{120}^o={60}^o$.

Hence, option $B$ is correct.