In the following figure,lines $AB$ and $CD$ intersect each other at point $P$. If $\angle APC : \angle BPC = 7 : 8$,find all the angles.

(N/A) Since $AB$ is a straight line,$\angle APC + \angle BPC = 180^{\circ}$ (Linear pair of angles).
Given that $\angle APC : \angle BPC = 7 : 8$.
Let $\angle APC = 7x$ and $\angle BPC = 8x$.
Then,$7x + 8x = 180^{\circ}$.
$15x = 180^{\circ}$.
$x = \frac{180^{\circ}}{15} = 12^{\circ}$.
Therefore,$\angle APC = 7 \times 12^{\circ} = 84^{\circ}$ and $\angle BPC = 8 \times 12^{\circ} = 96^{\circ}$.
Since $AB$ and $CD$ are straight lines intersecting at $P$,the vertically opposite angles are equal.
Thus,$\angle DPB = \angle APC = 84^{\circ}$ and $\angle APD = \angle BPC = 96^{\circ}$.
So,the four angles are $84^{\circ}, 96^{\circ}, 84^{\circ},$ and $96^{\circ}$.