In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on line CD such that DQ = 2 and QC = 3. Find sin angle PAQ.
AP = sqrt [5^2 + 4^2] = sqrt 41
PQ = sqrt [2^2 + 1^1] = sqrt 5
AQ = sqrt [ 5^2 + 2^2] = sqrt 29
Law of Cosines
PQ^2 = AP^2 + AQ^2 - 2 (AP * AQ) cos (PAQ)
5 = 41 + 29 - 2 [sqrt [41] *sqrt[29] cos (PAQ)
cos (PAQ) = [ 5 - 41 - 29 ] / [ - 2sqrt [41]sqrt[29] ] = 65 / sqrt [ 4756 ]
sin PAQ = sqrt [ 1 - 65^2 / 4756 ] = 3sqrt [ 70151] / 2378