Before I catch up on the 32 hours of sleep I have missed...
In right triangle ABC with angle B = 90 degrees, we have 2 sin A = 3 cos A. What is sin A?
Thanks!
2 sin A = 3 cos A square both sides
4 sin^2 A = 9 cos^2 A
4 sin^2 A = 9 [ 1 - sin^2 A]
4 sin ^2 A = 9 - 9 sin^2 A rearrange as
13 sin^2 A = 9
sin^2 A = 9 /13 sinc A is acute and < 90° ...take the positive root
sin A = 3 / √13 = ( 3 √13 ) / 13
Thanks CPhill!
+1452 
