In figure, $\angle A C B=40^{\circ}$. Find $\angle O A B$.
Given, $\angle A C B=40^{\circ}$
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
$\therefore \quad \angle A O B=2 \angle A C B$
$\Rightarrow \quad \angle A C B=\frac{\angle A O B}{2}$
$40^{\circ}=\frac{1}{2} \angle A O B$
$\Rightarrow \quad \angle A O B=80^{\circ}$ $\ldots$ (i)
In $\triangle A O B$, $A O=B O$ [both are the radius of a circle]
$\Rightarrow$ $\angle O B A=\angle O A B$ ... (ii)
[angles opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle $A O B$ is $180^{\circ}$.
$\therefore$ $\angle A O B+\angle O B A+\angle O A B=180^{\circ}$
$\Rightarrow$ $80^{\circ}+\angle O A B+\angle O A B=180^{\circ}$ [from Eqs. (i) and (ii)]
$\Rightarrow \quad 2 \angle O A B=180^{\circ}-80^{\circ}$
$\therefore$ $\angle O A B=\frac{100^{\circ}}{2}=50^{\circ}$