trig
posted by rick .
If sin = 4/5 with A in QII, and sin B = 3/5 with B in QIV find cos(AB)

cos(AB)= cosAcosB+sinAsinB
If sinA = 4/5 in QII
draw you triangle in QII, and find the third side.
The third side is 3. (Do the same thing for sin B= 3/5
Plug in:
cosAcosBsinAsinB
=(3/5)*(4/5)+(4/5)*(3/5)
=(12/25) (12/25)
=24/25 
53. the longest side is 55 cm. find the length of the shortest side