Question

In fig, $ABCD$ is a cyclic quadrilateral. if $\angle D={120}^0$, then $\angle CBE$ will be :

• A. ${120}^0$
• B. ${60}^0$
• C. ${180}^0$
• D. ${90}^0$

Answer 1

The correct option is A ${120}^0$

We know that opposite angles of cyclic quadrilateral are supplementary.

Then, $\angle ADC+\angle ABC={180}^o$

$⟹$ ${120}^o+\angle ABC={180}^o$

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

Also, $\angle ABC+\angle CBE={180}^o$ ...(Linear pair)

$⟹$ $\angle {60}^o+\angle CBE={180}^o$

$⟹$ $\angle CBE={180}^o-{60}^o={120}^o$

Thus, option $A$ is correct.