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!