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 $ABX$?

Guest Feb 11, 2020

#1**+1 **

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

CPhill Feb 12, 2020

#3**+1 **

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 _{}

Dragan Feb 12, 2020