Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If $BY = 2$ and $BZ = 4$, then what is the area of rectangle $WXYZ$?
W X
Z A B Y
4 2
Triangle ZWB is a 30-60-90 right triangle
Angle ZWB = 60
Angle WBZ = 30
So WZ = BZ/sqrt 3 = 4 /sqrt 3
Area of WXYZ = ZY * WZ = 6 * 4/sqrt 3 = 24/sqrt 3 = 8 sqrt 3
+595 
