Trig Problem

We know, $\sin(x)=\cos(\frac{\pi}{2}-x)$.

$\angle BAD = 90 = \frac{\pi}{2}$, exactly!

$\because \angle BAC = \angle BAP + \angle PAQ + \angle QAD = \frac{\pi}{2}$

$\therefore\angle BAP+\angle QAD=\frac{\pi}{2}-\angle PAQ$

$\therefore \sin(\angle PAQ)=\cos(\angle BAP + \angle QAD)=\cos(\angle BAP)\cos(\angle QAD)-\sin(\angle BAP)\sin(\angle QAD)$.

Triangle BAP and QAD are just right triangles, so it's easy to calculate trigonometric values off them.

We get $\frac{17\sqrt{1189}}{1189}$, as our final answer.