In triangle $ABC$, we have $\angle A = 30^\circ$, $\angle B = 60^\circ$, and $\angle C=90^\circ$. Point $X$ is on $\overline{AC}$ such that $\overline{BX}$ bisects $\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$?
+129852 See the following image :
AB is the hypotenuse of a 30-60-90 right triangle
Since BC is opposite the 30° angle it = (1/2)(12) = 6
And since AC is opposite the 60° angle it = 6√3
The area of triangle ABC = (1/2)(BC)(AC) = (1/2)(6)(6√3) = 18√3 (1)
Since BX is an angle bisector of angle B, then angle XBC = 30°
So triangle XBC is also a 30-60-90 right triangle
And angle CXB = 60°
And the side opposite this angle = BC = 60
So.....the side opposite angle XBC = 6/√3 = CX
So....the area of triangle XBC = (1/2) (product of the legs) = (1/2) BC * CX = (1/2) (6) (6/√3) =
18 / √ 3 = 6√3 (2)
So....the area of triangle ABX = (1) - (2) = 18√3 - 6√3 = 12√3 units^2
+1490 In triangle $ABC$, we have $\angle A = 30^\circ$, $\angle B = 60^\circ$, and $\angle C=90^\circ$. Point $X$ is on $\overline{AC}$ such that $\overline{BX}$ bisects $\angle ABC$. If $AB = 12$, then what is the area of triangle $ABX$?
Area of the triangle ABX = ( AB/2) * [( AB/2) * tan(A)]
Area ABX ≈ 20.785 u² 
