In triangle ABC, CA=4sqrt(2), CB=4sqrt(3), and A=\(45\) degrees. What is B in degrees?
+2667 Draw altitude \(CM\).
We know that \(\triangle ACM = 45-45-90\).
Thus, \(CM = AM = 4\).
In \(\triangle CBM\), we know that \(CB = 4\sqrt3\) and \(CM = 4\)
We can form the equation: \(\sin(B) = {\sqrt3 \over 3}\)
Taking the inverse, we find that \(\color{brown}\boxed {B \approx 35.264}\)
Here is the diagram: 
