+12 In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 4 and QC = 1. Find \(\sin \angle PAQ.\)
+128707 A B
4
P
1
D 4 Q 1 C
Let A = (0, 5)
P = (5 , 1)
Q = (4,0)
AP = sqrt [ 5^2 + 4^2 ] = sqrt (41)
AQ = sqrt [ 4^2 + 5^2 ] = sqrt (41)
PQ = sqrt [ 1^2 + 1^2) = sqrt (2)
Using the Law of Cosines
PQ^2 = AQ^2 + AP^2 - 2 ( AQ * AP) cos (PAQ)
2 = 41 + 41 - 2 ( sqrt 41 * sqrt 41) cos (PAQ)
[ 2 - 41 -41 ] / [ -2 * sqrt 41 * sqrt 41 ] = cos PAQ
-80 / [- 2 * 41] = cos PAQ
80 / 82 = 40 / 41 = cos PAQ
sin PAQ = sqrt [ 41^2 - 40^2 ] / 41 = sqrt (81) / 41 = 9 / 41
