+285 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.
+129845 A B
4
P
1
D 1 Q 4 C
AB = 5
AP^2 = 5^2 + 4^2 = 41
AD = 5
AQ^2 = 5^2 + 1^2 = 26
PQ^2 = 4^2 + 1^2 = 17
Law of Cosines
PQ^2 = AP^2 + AQ^2 - 2 (AP* AQ) cos (PAQ)
[PQ^2 - AP^2 - AQ^2 ] / [ -2 sqrt (AP * AQ) ] = cos PAQ
[ 17 - 41 - 26 ] / [ -2 sqrt (26 * 41) ] = cos PAQ = 25 / sqrt 1066 = cos PAQ
sin PAQ = sqrt [ 1066 - 25^2 ] / sqrt 1066 = sqrt (441) / sqrt 1066 = 21 / sqrt 1066
