+1248 Points $A$ and $B$ are on side $\overline{YZ}$ of rectangle $WXYZ$ such that $\overline{WA}$ and $\overline{WB}$ trisect $\angle ZWX$. If the area of triangle $WAZ$ is 6, what is the area of triangle $WAB$?
+129771 Triangle WAZ is a 30-60-90 right triangle
Angle AWZ = 30
So angle BWZ = 60
WZ / AZ = sqrt 3
WZ = sqrt 3 AZ
Triangle BWZ is a 30 - 60 -90 right triangle
BZ = sqrt (3) WZ
Area of WAZ = (1/2) AZ * WZ
6 = (1/2) AZ * WZ
12 = AZ * WZ
So area of triangle BWZ = (1/2) BZ *WZ = (1/2) [sqrt (3) * WZ] * [sqrt (3) AZ ] =
(1/2) (sqrt 3)^2 * AZ * WZ =
(3/2) * 12 = 18
[ WAB ] = [ BWZ ] - [WAZ] = 18 - 6 = 12
