+1911 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 $BXA$.
+129850 A
X
B 12 C
AB = 12sqrt 3
AC = 24
sin (angle A) = sin (30) = (1/2)
Angle ABX = 45
Angle BXA = 180 - 30 - 45 = 105
Law of Sines
BX /sin (30) = AB / sin (105)
BX / sin (30) = 12 sqrt 3 / sin (105)
BX = 12sqrt (3) sin (30) /sin (105) = 12sqrt (6) / (sqrt (3) + 1)
[BXA ] = (1/2)AB*BX *sin (45) = (1/2) (12sqrt 3)(12sqrt6) / (sqrt (3)+ 1) *sqrt (2) / 2 =
216 / (1 + sqrt 3) = 216 ( 1 -sqrt 3) / [ 1 - 3] = 216 [ sqrt (3) - 1] / 2 = 108 (sqrt (3) - 1)
