In the below figure, A, B, and C are three points on a circle with center O such that $\angle BOC={{30}^{\circ }}$ and $\angle AOB={{60}^{\circ }}$. If D is a point on the circle other than the arc ABC, find $\angle ADC$.

Hint: We are going to use the theorem which states that the angle formed by the arc on the center of the circle is the double of the angle subtended at any point on the remaining part of the circle. For that, we are going to, first of all, add angle AOB and angle BOC and then apply this theorem.

Complete step by step answer:
The figure given in the above problem is as follows:

We are asked to find the angle ADC $\left( \angle ADC \right)$ which we are going to find by using the theorem which says that angle subtended by an arc on the center of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.
From this theorem, we can say that angle AOC is double that of angle ADC. Writing what we have just said in the mathematical form we get,
$\angle AOC=2\angle ADC$
Now, angle AOC is the sum of angle AOB and angle BOC so writing angle AOC in terms of angle AOB and angle BOC we get,
$\angle AOB+\angle BOC=2\angle ADC$
It is given that $\angle AOB={{60}^{\circ }}\And \angle BOC={{30}^{\circ }}$ so using these values in the above equation we get,
$\begin{align}
  & {{60}^{\circ }}+{{30}^{\circ }}=2\angle ADC \\
 & \Rightarrow {{90}^{\circ }}=2\angle ADC \\
\end{align}$
Dividing 2 on both the sides of the above equation we get,
$\begin{align}
  & \dfrac{{{90}^{\circ }}}{2}=\angle ADC \\
 & \Rightarrow {{45}^{\circ }}=\angle ADC \\
\end{align}$
From the above solution, we got the value of angle ADC as ${{45}^{\circ }}$.

Note: To solve the above problem you must know the theorem which says that the angle formed by the arc on the center of the circle is the double of the angle subtended at any point on the remaining part of the circle. If you don’t know this theorem then you cannot solve this problem.