Let $O$ be the intersection of $\overline{AP}$ and $\overline{DQ}$.
[asy] unitsize(0.5cm); pointpen=black; pointfontpen=fontsize(9pt); pair A=(0,0), B=(0,5), C=(5,5), D=(5,0), P=(0,4), Q=(4,5), O=(5,4), N=(5,5); path DQ=D--Q, AP=A--P, PQ=P--Q; D(MP("A",A)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D)--cycle); D(MP("P",P,W) --MP("O",O,E)); D(D--MP("Q",Q,N) , gray+1.3); D(N--P, gray+1.3); [/asy]
By AA similarity, $\triangle DCP \sim \triangle BAP$. Thus
\[\frac{DC}{AP} = \frac{CP}{BP} \quad \Rightarrow \quad AP = \frac54\]and \[DC = \frac54 \cdot 5 = \frac{25}{4}.\]
By the Pythagorean Theorem applied to right triangle $DQC$, we find
\[DC = \sqrt{DQ^2 + QC^2} = 5.\]
By the Law of Cosines applied to triangle $DCP$, we get
\begin{align*} DP^2 &= DC^2 + CP^2 - 2 \cdot CP \cdot DC \cdot \cos \angle PDC \\ &= 5^2 + 1^2 - 2 \cdot 5 \cdot 1 \cdot \cos \angle PDC \\ &= 25 + 1 - 10 \cos \angle PDC . \end{align*}
Similarly, by the Law of Cosines, $AP^2 = \left(\dfrac54\right)^2 + 4^2 - 2 \cdot \dfrac54\cdot 4\cdot \cos \angle DAP.$
Adding these two results gives us
\[DP^2 + AP^2 = 26 - 10 \cos \angle PDC - 2 \cdot \dfrac54 \cdot 4 \cdot \cos \angle DAP.\]
By the Law of Cosines applied to triangle $DAP$, we know
$\cos \angle DAP = -\cos \angle PDC,$ so $2 \cdot \dfrac54\cdot 4 \cdot \cos \angle DAP = 10 \cos \angle PDC.$
It follows that $DP^2 + AP^2 = 26.$
Finally, the Law of Cosines on triangle $APQ$ gives us
\begin{align*} AQ^2 &= AP^2 + DP^2 - 2DP\cdot AP \cdot \cos \angle PAQ \\ &= 26 - 2 \cdot \dfrac54\cdot 4 \cdot \cos \angle PAQ \\ &= 26 - 10 \cos \angle PAQ . \end{align*}
Since $AP > 0,$ we must have $DP > 0,$ so $DP^2 > 0,$ and
\[\cos \angle PAQ = \frac{26 - AQ^2}{10}.\]
By the Extended Law of Sines ($\left(\dfrac{ABC}{2R}\right) = \sin A = \sin B = \sin C$), $\sin \theta = \sin (\pi - \theta) = \sin \angle PAQ$ when $\angle PAQ < 180^\circ.$
We compute $AQ^2 = 26 - 10 \cos \angle PAQ.$ Therefore,
\[\sin \angle PAQ = \sin (\pi - \angle PAQ) = \sin \angle PAQ = \sin \theta = \sqrt{1 - \cos^2 \theta} = \boxed{\frac{3}{5}}.\]
In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 1 and QC = 4. Find sin angle PAQ.
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