In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on CD such that DQ= 4 and QC = 1. Find sin angle PAQ
Let's label the points as follows:
[asy]
unitsize(0.5cm);
pair A, B, C, D, P, Q;
A = (0,0);
B = (0,6);
C = (6,6);
D = (6,0);
P = (0,4);
Q = (6,4);
draw(A--B--C--D--A);
draw(P--A--Q);
draw(P--C,dotted);
draw(Q--D, dotted);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
label("$P$", P, W);
label("$Q$", Q, E);
[/asy]
We have BP = 4 and PC = 1, so by the Pythagorean theorem, $BC = \sqrt{BP^2 + PC^2} = \sqrt{4^2 + 1^2} = \sqrt{17}$. Also, DQ = 4 and QC = 1, so $CQ = BC - BQ = \sqrt{17} - 4$. Therefore, AQ = AD + DQ = $\sqrt{17} + 4$.
Since triangle APQ is right, we can use the sine function to find $\sin \angle PAQ$:$$\sin \angle PAQ = \frac{AQ}{AP} = \frac{\sqrt{17} + 4}{\sqrt{17}} = \boxed{\frac{4}{\sqrt{17}}}.$$
To find sin angle PAQ, we need to determine the lengths of PA and AQ.
Considering triangle BPC, we can use the Pythagorean Theorem:
BP^2 + PC^2 = BC^2
4^2 + 1^2 = BC^2
16 + 1 = BC^2
BC^2 = 17
BC = sqrt(17)
Now, let's look at triangle DQC:
DQ^2 + QC^2 = DC^2
4^2 + 1^2 = DC^2
16 + 1 = DC^2
DC^2 = 17
DC = sqrt(17)
Since ABCD is a square, BC = CD. Thus, BC = CD = sqrt(17).
Now, let's analyze triangle APD. We know that AD = DP, as they are sides of a square. Therefore, APD is an isosceles triangle.
Since triangle APD is isosceles, we can conclude that angle PAD = angle PDA.
Considering triangle PAQ, we know that angle PAQ = angle PAD + angle DAQ.
Since angle PAD = angle PDA, we have:
angle PAQ = angle PAD + angle DAQ
angle PAQ = angle PDA + angle DAQ
If we know the values of angle PDA and angle DAQ, we can find angle PAQ.
However, we need more information to determine the lengths of PA and AQ or the angles PDA and DAQ to calculate sin angle PAQ.