Question

In the figure above $ABCD$ is a cyclic quadrilateral, $O$ is the centre of the circle. If $\angle OAD={30}^{\circ }$ and $\angle OCD={20}^{\circ }$ then find $\angle ABC$.

• A. ${120}^{\circ }$
• B. ${50}^{\circ }$
• C. ${130}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is C ${130}^{\circ }$

$\angle ODA=\angle OAD$ ..... (Radius is equal so corresponding angles are equal)

Similarly,

$\angle ODC=\angle OCD$

$\therefore \angle ADC={30}^o+{20}^o={50}^o$

$\angle ABC=180-\angle ADC$ ..... (Property of cyclic quadrilateral)

$\angle ABC={130}^o$

Hence, ${130}^o$ is the correct answer.