In figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.
Given, $\angle O A B=30^{\circ}$ and $\angle O C B=57^{\circ}$
In $\triangle A O B$. $A O=O B$ [both are the radius of a circle]
$\Rightarrow \quad \angle O B A=\angle B A O=30^{\circ}$ [angles opposite to equal sides are equal]
In $\triangle A O B$,
$\Rightarrow$ $\angle A O B+\angle O B A+\angle B A O=180^{\circ}$ [by angle sum property of a triangle]
$\therefore \quad \angle A O B+30^{\circ}+30^{\circ}=180^{\circ}$
$\therefore$ $\angle A O B=180^{\circ}-2\left(30^{\circ}\right)$
$=180^{\circ}-60^{\circ}=120^{\circ}$ $\ldots(\mathrm{i})$
Now, in $\triangle O C B$,
$O C=O B$ [both are the radius of a circle]
$\Rightarrow \quad \angle O B C=\angle O C B=57^{\circ}$
[angles opposite to equal sides are equal]
In $\triangle O C B$
$\angle C O B+\angle O C B+\angle C B O=180^{\circ}$ [by angle sum property of triangle]
$\therefore$ $\angle C O B=180^{\circ}-(\angle O C B+\angle O B C)$
$=180^{\circ}-\left(57^{\circ}+57^{\circ}\right)$
$=180-114^{\circ}=66^{\circ}$ ...(ii)
From Eq. (i), $\angle A O B=120^{\circ}$
$\Rightarrow \quad \angle A O C+\angle C O B=120^{\circ}$
$\Rightarrow \quad \angle A O C+66^{\circ}=120^{\circ} \quad$ [from Eq. (ii)]
$\therefore \quad \angle A O C=120^{\circ}-66^{\circ}=54^{\circ}$