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

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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?

  • $\begingroup$ Clever method, but a little more explanations might be needed for the OP to comprehend. $\endgroup$ – Szeto Jul 8 at 5:52
  • 2
    $\begingroup$ 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. $\endgroup$ – Rodrigo Pizarro Jul 8 at 5:56

Alternatively, from the Sine theorem: $$\frac{PQ}{\sin x}=\frac{BQ}{\sin \angle BPQ} \Rightarrow\\ \sin \angle BPQ=\frac{BQ\sin x}{PQ}=\frac{CQ}{PQ}=\frac12 \Rightarrow \angle BPQ=30^\circ \Rightarrow \angle PAQ=15^\circ \Rightarrow \\ 2x=90-\angle PAQ=75^\circ \Rightarrow x=37.5^\circ.$$


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