+0 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 $BXA$.
+1632 Triangle ABC is a 30-60-90 with base BC and height AB. BC = 12, so AB = 12sqrt(3), and thus, the hypotenus AC = 24.
By angle bisector theorem, CX/AX = 12/[12sqrt(3)] = 1/sqrt(3), cross multiplying you get CXsqrt(3) = AX.
Additionally, CX + AX = AC = 24. By substitution, CX + CXsqrt(3) = 24, CX = 24/(1 + sqrt(3)) = [24sqrt(3) - 24]/(3 - 1) after rationalization of the denominator, which simplifies to 12sqrt(3) - 12 = CX. AX = CXsqrt(3) = 36 - 12sqrt(3).
Now for some trig: The area of a triangle with side lengths a b c, aside from base * height / 2, is also written as ab*sinC/2, where C is the angle included between sides a and b (you should remember this). In our problem, specifically, [BXA] (this denotes the area of triangle BXA) = AB*AX*sin(BAX)/2 = 12sqrt(3)*(36 - 12sqrt(3))*sin(30 degrees)/2 = (432sqrt(3) - 432)/4 = 108sqrt(3) - 108 = [BXA]
+1632 Triangle ABC is a 30-60-90 with base BC and height AB. BC = 12, so AB = 12sqrt(3), and thus, the hypotenus AC = 24.
By angle bisector theorem, CX/AX = 12/[12sqrt(3)] = 1/sqrt(3), cross multiplying you get CXsqrt(3) = AX.
Additionally, CX + AX = AC = 24. By substitution, CX + CXsqrt(3) = 24, CX = 24/(1 + sqrt(3)) = [24sqrt(3) - 24]/(3 - 1) after rationalization of the denominator, which simplifies to 12sqrt(3) - 12 = CX. AX = CXsqrt(3) = 36 - 12sqrt(3).
Now for some trig: The area of a triangle with side lengths a b c, aside from base * height / 2, is also written as ab*sinC/2, where C is the angle included between sides a and b (you should remember this). In our problem, specifically, [BXA] (this denotes the area of triangle BXA) = AB*AX*sin(BAX)/2 = 12sqrt(3)*(36 - 12sqrt(3))*sin(30 degrees)/2 = (432sqrt(3) - 432)/4 = 108sqrt(3) - 108 = [BXA]
