, 2016

Since points $Y$, $W$ and $Q$ form a straight line segment, then $\angle YWV = 180^\circ - \angle VWQ$ and so $\angle YWV = 180^\circ - 125^\circ = 55^\circ$. Since $Q'$ is the final position of $Q$ after folding, then $\angle Q'WV$ is the final position of $\angle QWV$ after folding, and so $\angle Q'WV = \angle QWV$. Thus, $\angle Q'WV = \angle QWV = 125^\circ$ and so $\angle Q'WY = \angle Q'WV - \angle YWV = 125^\circ - 55^\circ = 70^\circ$. [[IMAGE0]] Since $Q'W$ and $R'Y$ are parallel sides of the piece of paper, then $\angle R'YW + \angle Q'WY = 180^\circ$, and so $\angle R'YW = 180^\circ - \angle Q'WY = 180^\circ - 70^\circ = 110^\circ$. Finally, $\angle PYV$ is opposite $\angle R'YW$ so $\angle PYV = \angle R'YW = 110^\circ$.