What is the measure of the PQ segment in the figure below? Given a semi-circumference of diameter AB and center «O» and
let P and Q be points on this curve, we draw $QH\perp OB (H \in  OB)$ .
Calculate "PQ", if AP = 2; QH = 6, $ m\measuredangle PAQ=3m\measuredangle QAB$
My progress:(hard question)

$\triangle BAQ: 6^2 = m.n\rightarrow 36 = mn\\
6q = a.b\\
 \measuredangle QPB = \measuredangle QAB = x\\
 \measuredangle PBQ = \measuredangle PAQ = 3x\\
 \measuredangle ABP = \measuredangle AQP = \alpha\\
\measuredangle Q + \measuredangle A = 180^o\implies \boxed{4x+\alpha =90^o}\therefore \measuredangle A = 90-\alpha\\
\measuredangle B = 3x+\alpha = 3x+90-4x = 90^o -x\\
T.Ptolomeu: b.c =e.q+2a\\
\triangle BPA: c^2+4 = q^2\rightarrow q^2 - c^2 = 4 \\
\triangle QHA: b^2 = 36+n^2\rightarrow b^2-n^2= 36$

Here is a solution using trigonometry. Say radius of the circle is $r$ and $O$ is the center of the circle.
$\angle BQH = x, \angle QOB = 2x$
So, $a = \frac{6}{\cos x}$
$\frac{a}{2} = OB \sin x \implies r = \frac{6}{\sin 2x}$
But in $\triangle ABP, AB = 2 r = \frac{2}{\cos 4x}$
Equating, $6 \cos 4x = \sin 2x$
$6 - 12 \sin^2 2x = \sin 2x$
Solving the quadratic, $\sin 2x = \frac{2}{3}$
$\cos 4x = \frac{1}{9}, r = 9$
$\cos 2x = \frac{\sqrt5}{3}, \cos x = \frac{\sqrt5 + 1}{2 \sqrt3}, \sin x = \frac{\sqrt5 - 1}{2 \sqrt3}$
So, $\sin 3x = \frac{7 + \sqrt5}{6 \sqrt3}$
$ \small PQ = 2 r \sin 3x = 7 \sqrt3 +  \sqrt{15} \approx 16$
