+1459 In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 2 and QC = 3. Find sin angle PAQ.
+399 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.
