+58 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.
Using the Law of Cosines
BC^2 = DB^2 + DC^2 - 2(DB*DC) cos (60 degrees)
BC^2 = 2^2 + 5^2 - 2 (2*5)(1/2)
BC^2 = 29 - 10
BC = sqrt (19)
Using the Law of Sines
sin BCD / BD = sin BDC / BC
sin BCD / 2 = sqrt (3) / [ 2 sqrt (19) ]
sin BCD = sqrt (3/19)
BCD = arcsin (3/19) ≈ 23.4°
Angle ACE = 90 - 23.4 ≈ 66.6°
Angle CAE = 180 -60 - 66.6 ≈ 53.4°
Using the Law of Sines again
CE / sin CAE = AE /sin ACE
2 / sin (53.4) = AE / sin (66.6)
AE = 2sin (66.6) / sin (53.4) ≈ 2.286
+129852 
Since PC and QC = 1, then PQ = sqrt (2)
Let A = (0,,0) and P = (4,5) so AP =sqrt [ 4^2 + 5^2] = sqrt (41) = AQ
So....by Law of Cosines
PQ^2 = AP^2 + AQ^2 -2 (AP)(AQ) 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 [ 1 - (40/41)^2 ] = sqrt [ 41^2 - 40^2] / 41 = sqrt (81) / 41 = 9 / 41
CORRECTED ANSWER !!!!
+58 Thanks for the answer, it helped me a lot!
5sqrt(21)/41 wasn't correct, but your response helped me figure out what to do next.
I followed your solution and got mostly the same results except that I got 40/41 as cos(PAQ) instead of 34/41.
I plugged that into the relationship between the sine and cosine of PAQ and got 9/41 as the final answer, which turned out to be correct.
\( sin(PAQ) = \boxed{9/41}\)
