+1036 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 $BXA$.
+128578 30 A
X
90 B 12 C
Angle B = 60
AB = 12sqrt 3
AC = 24
Since BX is a bisector...let XC = 24-m and AX = m
And we have that
m / AB = (24-m) / BC
m / (sqrt 3) = (24 -m)
m = (24 - m) sqrt 3
m = 24sqrt 3 - msqrt 3
m+ msqrt 3 = 24sqrt 3
m ( 1 + sqrt 3) =24sqrt 3
m = 24sqrt 3 / ( 1 + sqrt 3) = 12sqrt3 ( sqrt 3 - 1) = 12 [ 3 - sqrt 3 ]
[ BXA ] = (1/2) AX * AB * sin 30° = (1/4) AX * AB = 12 [3 -sqrt 3 ] * [12 sqrt 3] / 4 =
3 [ 3 -sqrt 3 ] [ 12sqrt 3 ] = 108 (sqrt 3 - 1)
