Question

$ABCD$ is a cyclic quadrilateral inscribed in a circle with the centre $O$. Then $\angle OAD$ is equal to:

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (D)

${60}^{\circ }$

In $△OAB$,
$OA=OB$ (radius of circle).
Thus, $\angle OAB=\angle OBA={40}^o$...[Isosceles triangle property].

Similarly, $\angle OBC=\angle OCB={30}^o$
and, $\angle OCD=\angle ODC={50}^o$
and, $\angle ODA=\angle OAD=x$.

Sum of angles of the quadrilateral $={360}^o$.
Then, $\angle OAB+\angle OBA+\angle OBC+\angle OCB+\angle OCD+\angle ODC+\angle ODA+\angle OAD={360}^o$
$⟹$ $2({40}^o+{30}^o+{50}^o+x)={360}^o$
$⟹$ ${120}^o+x={180}^o$
$⟹$ $x={60}^{\circ }$.

Thus, $\angle OAD={60}^{\circ }$.

Hence, option $D$ is correct.