+1443 In square ABCD, P is on BC such that BP = 5 and PC = 2, and Q is on CD such that DQ = 4 and QC = 3. Find sin angle PAQ.
+129850 A B
5
P
2
D 4 Q 3 C
Let A = (0,7)
Let P = (7,2)
Let Q = (4,0)
AP = sqrt ( 7^2 + 5^2) = sqrt ( 74)
AQ = sqrt ( 4^2 + 7^2) = sqrt (65)
PQ = sqrt (3^2 + 2^2) = sqrt (13)
Law of Cosines
PQ^2 = AP^2 + AQ^2 - 2 ( sqrt [AP * AQ ]) cosPAQ
13 = 74 + 65 - 2 sqrt (74 * 65) cos PAQ
cos PAQ = [ 13 - 74 - 65 ] / [ -2 *sqrt( 74 *65) ] = 63 / sqrt (4810)
sin PAQ = sqrt [4810 - 63^2 ] / sqrt [ 4810 ] = sqrt [841 / 4810 ]
