Prove the following

Solution:

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)]

$\therefore \quad \angle A O C=120^{\circ}-66^{\circ}=54^{\circ}$