In the figure,$O$ is the centre of the circle,and $\angle BCO = 30^{\circ}$. Find $x$ and $y$.

(N/A) $O$ is the centre of the circle and $\angle BCO = 30^{\circ}$. We need to find the values of $x$ and $y$.
In the right-angled triangle $\Delta OPC$,we have:
$\angle POC = 180^{\circ} - (\angle OPC + \angle PCO)$
$\Rightarrow \angle POC = 180^{\circ} - (90^{\circ} + 30^{\circ}) = 60^{\circ}$.
Given that $\angle AOD = 90^{\circ}$,and since $AD$ is a straight line,$\angle AOD + \angle DOP = 180^{\circ}$ (linear pair).
$\angle DOP = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Now,$\angle COD = \angle DOP - \angle POC = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
$\angle CBD = \frac{1}{2} \angle COD = \frac{1}{2} \times 30^{\circ} = 15^{\circ}$. Thus,$y = 15^{\circ}$.
Also,$\angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.
In $\Delta ABP$,the sum of angles is $180^{\circ}$:
$x + (\angle ABD + y) + \angle APB = 180^{\circ}$
$x + (45^{\circ} + 15^{\circ}) + 90^{\circ} = 180^{\circ}$
$x + 60^{\circ} + 90^{\circ} = 180^{\circ}$
$x = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,$x = 30^{\circ}$ and $y = 15^{\circ}$.