In the figure,$\angle OAB = 30^{\circ}$ and $\angle OCB = 57^{\circ}$. Find $\angle BOC$ and $\angle AOC$.

(N/A) In $\Delta OBC$,we have:
$OB = OC$ (Radii of the same circle)
Therefore,$\angle OCB = \angle OBC = 57^{\circ}$ (Since $\angle OCB = 57^{\circ}$ is given,and angles opposite to equal sides are equal).
Now,in $\Delta BOC$,the sum of angles is $180^{\circ}$:
$\angle OCB + \angle OBC + \angle BOC = 180^{\circ}$
$57^{\circ} + 57^{\circ} + \angle BOC = 180^{\circ}$
$114^{\circ} + \angle BOC = 180^{\circ}$
$\angle BOC = 180^{\circ} - 114^{\circ} = 66^{\circ}$.
In $\Delta OAB$,we have:
$OA = OB$ (Radii of the same circle)
Therefore,$\angle OBA = \angle OAB = 30^{\circ}$.
In $\Delta OAB$,the sum of angles is $180^{\circ}$:
$\angle OAB + \angle OBA + \angle AOB = 180^{\circ}$
$30^{\circ} + 30^{\circ} + \angle AOB = 180^{\circ}$
$\angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Since $\angle AOB = \angle AOC + \angle BOC$,we have:
$120^{\circ} = \angle AOC + 66^{\circ}$
$\angle AOC = 120^{\circ} - 66^{\circ} = 54^{\circ}$.
Hence,$\angle BOC = 66^{\circ}$ and $\angle AOC = 54^{\circ}$.