In the figure,lines $PQ$ and $RS$ intersect each other at point $O$. If $\angle POR : \angle ROQ = 5 : 7,$ find all the angles.

(N/A) Since $PQ$ is a straight line,the sum of angles forming a linear pair is $180^o$.
$\angle POR + \angle ROQ = 180^o$ (Linear pair axiom)
Given that $\angle POR : \angle ROQ = 5 : 7$.
Let $\angle POR = 5x$ and $\angle ROQ = 7x$.
Substituting these into the equation:
$5x + 7x = 180^o$
$12x = 180^o$
$x = 15^o$
Therefore,$\angle POR = 5 \times 15^o = 75^o$ and $\angle ROQ = 7 \times 15^o = 105^o$.
Since $PQ$ and $RS$ are intersecting lines,vertically opposite angles are equal:
$\angle POS = \angle ROQ = 105^o$
$\angle SOQ = \angle POR = 75^o$
Thus,the angles are $75^o, 105^o, 75^o,$ and $105^o$.