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.

Answer 1

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
$$\frac{DC}{AP}=\frac{CP}{BP}\Rightarrow AP=\frac{5}{4}$$and $$DC=\frac{5}{4}\cdot 5=\frac{25}{4}.$$
By the Pythagorean Theorem applied to right triangle $DQC$, we find
$$DC=\sqrt{D{Q}^2+Q{C}^2}=5.$$
By the Law of Cosines applied to triangle $DCP$, we get
$$\begin{array}{rl}D{P}^2& =D{C}^2+C{P}^2-2\cdot CP\cdot DC\cdot \cos \mathrm{\angle }PDC\\ & =5^2+1^2-2\cdot 5\cdot 1\cdot \cos \mathrm{\angle }PDC\\ & =25+1-10\cos \mathrm{\angle }PDC.\end{array}$$
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
$$D{P}^2+A{P}^2=26-10\cos \mathrm{\angle }PDC-2\cdot {\displaystyle \frac{5}{4}}\cdot 4\cdot \cos \mathrm{\angle }DAP.$$
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
$$\begin{array}{rl}A{Q}^2& =A{P}^2+D{P}^2-2DP\cdot AP\cdot \cos \mathrm{\angle }PAQ\\ & =26-2\cdot {\displaystyle \frac{5}{4}}\cdot 4\cdot \cos \mathrm{\angle }PAQ\\ & =26-10\cos \mathrm{\angle }PAQ.\end{array}$$
Since $AP > 0,$ we must have $DP > 0,$ so $DP^2 > 0,$ and
$$\cos \mathrm{\angle }PAQ=\frac{26-A{Q}^2}{10}.$$
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,
$$\sin \mathrm{\angle }PAQ=\sin (\pi -\mathrm{\angle }PAQ)=\sin \mathrm{\angle }PAQ=\sin \theta =\sqrt{1-{\cos }^2\theta }=\boxed{{\displaystyle \frac{3}{5}}}.$$