Solve for the angle $x$ in the right triangle without trigonometry.
I don't know how to find the angle $x$.
I drew the height from $P$ to $AQ$, because the triangle $APQ$ is isosceles. Also i noticed that drawing a perpendicular from $P$ to $BC$ may be useful, because it is also a bisector of $\angle QPB$. After that i tried some similarity between triangles, but found nothing. Any hints?
PS: I got this problem from an exercise list, and found that they use the angle approximation of $37,53,90$ in the triangle $3-4-5$ sometimes. I don't know if this is the case.