In triangle $ABC$, $\angle A = 30^\circ$ and $\angle B = 90^\circ$. Point $X$ is on side $\overline{AC}$ such that line segment $\overline{BX}$ bisects $\angle ABC$. If $BC = 12$, then find the area of triangle $BXA$.
A
30
X
B 90 C 60
12
AB = 12sqrt 3
Law of Sines
AB / sin BXA = BX / sin BXA
12sqrt 3 / sin 75 = BX / sin 30
BX = 6sqrt 3 / sin 75
[ ABX ] = (1/2) BX * AB * sin (45) =
(1/2) (6sqrt 3 / sin 75) ( 12sqrt 3) sin (45)
108 sin (45) / sin(75) =
216 / (1 + sqrt 3) ≈ 79.06
+269 
