ABCD is a cyclic quadrilateral inscribed in a circle with the center O. Find $\angle \mathrm{OAD}$.

Consider the given figure,

Given-
ABCD is a cyclic quadrilateral.
$\angle \mathrm{OAB}={40}^{\mathrm{o}},\angle \mathrm{OBC}={30}^{\mathrm{o}}\mathrm{ }\angle \mathrm{OCD}={50}^{\mathrm{o}}$
$\mathrm{OC}=\mathrm{OB}$                                       (Radius of the circle)
$\therefore \mathrm{\Delta OBC}$ is an isosceles triangle.
Then,$\angle \mathrm{OCB}=\angle \mathrm{OBC}={30}^{\mathrm{o}}$
From figure,
$\angle \mathrm{BCD}=\angle \mathrm{OCD}+\angle \mathrm{OCB}$
$\angle \mathrm{BCD}={50}^{\mathrm{o}}+{30}^{\mathrm{o}}$
$\angle \mathrm{BCD}={80}^{\mathrm{o}}$
We know that,
In a cyclic quadrilateral sum of opposite angles are supplementary.
Then,
$\angle \mathrm{BAD}+\angle \mathrm{BCD}={180}^{\mathrm{o}}$
$\Rightarrow \angle \mathrm{BAD}={180}^{\mathrm{o}}-\angle \mathrm{BCD}$
$\Rightarrow \angle \mathrm{BAD}={180}^{\mathrm{o}}-{80}^{\mathrm{o}}$
$\Rightarrow \angle \mathrm{BAD}={100}^{\mathrm{o}}$
From figure,
$\mathrm{\angle }\mathrm{BAD}=\angle \mathrm{OAB}+\angle \mathrm{OAD}$
$\Rightarrow \angle \mathrm{OAD}=\angle \mathrm{BAD}-\angle \mathrm{OAB}$
$\Rightarrow \angle \mathrm{OAD}={100}^{\mathrm{o}}-{40}^{\mathrm{o}}$
$\Rightarrow \angle \mathrm{OAD}={60}^{\mathrm{o}}$
Hence, the measure of $\angle \mathrm{OAD}$ is ${60}^{\mathrm{o}}$.
Therefore, option 60^o is correct.