In triangle $ABC$, $\angle A = 30^\circ$ and $\angle B = 60^\circ$. Point $X$ is on side $\overline{AC}$ such that line segment $\overline{BX}$ bisects $\angle ABC$. If $AC = 12$, then find the area of triangle $BXA$.
AC=12 not AB!!
+129885 30 A
12 X
4
90 C B 60
The triangle is 30 - 60 - 90
AC = 12 BC = 12/ sqrt (3) = 4sqrt (3) AB = 8 sqrt (3)
Area of triangle ABC = (1/2) BC * AC = (1/2) (4sqrt (3) * 12 = 24 sqrt (3)
XC = 4
Area of triangle BXC = (1/2) ( BC) (XC) = (1/2)(4sqrt (3)) ( 4) = 8 sqrt (3)
Area of triangle BXA = [24sqrt (3) - 8 sqrt (3) ] = 16sqrt (3)
Since ∠A=30∘ and ∠B=60∘, then ∠C=90∘. Also, ∠BXA=21⋅90∘=45∘.
Triangle ABX is a 45-45-90 triangle, so BX=AX=2212=32. Therefore, the area of triangle BXA is [\frac{1}{2} \cdot BX \cdot AX = \frac{1}{2} \cdot 3\sqrt{2} \cdot 3\sqrt{2} = \boxed{9}.]
+129885 30 A
12 X
4
90 C B 60
The triangle is 30 - 60 - 90
AC = 12 BC = 12/ sqrt (3) = 4sqrt (3) AB = 8 sqrt (3)
Area of triangle ABC = (1/2) BC * AC = (1/2) (4sqrt (3) * 12 = 24 sqrt (3)
XC = 4
Area of triangle BXC = (1/2) ( BC) (XC) = (1/2)(4sqrt (3)) ( 4) = 8 sqrt (3)
Area of triangle BXA = [24sqrt (3) - 8 sqrt (3) ] = 16sqrt (3)
