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 $\sin \angle PAQ$:$$\sin \angle PAQ = \frac{AQ}{AP} = \frac{\sqrt{17} + 4}{\sqrt{17}} = \boxed{\frac{4}{\sqrt{17}}}.$$
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
1 answer