# Solve for the angle $x$ in the right triangle without trigonometry.

Solve for the angle $$x$$ in the right triangle without trigonometry. I don't know how to find the angle $$x$$.

My try

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.

Let $$QM$$ be an altitude of $$\Delta PQB$$.
Thus, $$MQ=QC=\frac{1}{2}PQ,$$ $$\measuredangle MPQ=30^{\circ},$$ $$\measuredangle PAQ=15^{\circ}.$$ Can you end it now?
• I can comprehend. Actually in my drawings i drew that altitude, but i didn't notice the bisector rule to 2 perpendicular segments before. After that we see the beautiful $30-60-90$ triangle. – Rodrigo Pizarro Jul 8 at 5:56
Alternatively, from the Sine theorem: $$\frac{PQ}{\sin x}=\frac{BQ}{\sin \angle BPQ} \Rightarrow\\ \sin \angle BPQ=\frac{BQ\sin x}{PQ}=\frac{CQ}{PQ}=\frac12 \Rightarrow \angle BPQ=30^\circ \Rightarrow \angle PAQ=15^\circ \Rightarrow \\ 2x=90-\angle PAQ=75^\circ \Rightarrow x=37.5^\circ.$$