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.
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,
\[\sin \angle PAQ = \sin (\pi - \angle PAQ) = \sin \angle PAQ = \sin \theta = \sqrt{1 - \cos^2 \theta} = \boxed{\frac{3}{5}}.\]
To find sin angle PAQ, we can start by finding the lengths of the sides of triangle PAQ.
First, we can find the length of side AQ. Using the fact that DQ = 1 and QC = 4, we can infer that the length of AD is also 1 (since ABCD is a square). Therefore, the length of AQ is AD + DQ, which is 1 + 1 = 2.
Next, we can find the length of side PA. Since BP = 4 and PC = 1, we can find the length of BC by summing these two lengths: BC = BP + PC = 4 + 1 = 5. Since ABCD is a square, the length of BC is equal to the length of AD. Therefore, the length of PA is AD - PD, which is 1 - 4 = -3.
Now, we can use the lengths of sides AQ and PA to find sin angle PAQ. Recall that sin angle PAQ is given by the ratio of the opposite side (AQ) to the hypotenuse (PA):
sin angle PAQ = AQ / PA
Substituting the values we obtained:
sin angle PAQ = 2 / -3 = -2/3
Therefore, the value of sin angle PAQ is -2/3.