I really need help with this problem....I have spent over 3 hours on this one...

Since angle A is 30 degrees and angle B is 60 degrees, then angle C is 90 degrees. Therefore, triangle ABC is a 30-60-90 triangle.

The ratio of the sides of a 30-60-90 triangle is 1:sqrt(3):2. Since AC is 12, then AB is 12/2 = 6 and BC is 12sqrt(3)/2 = 6sqrt(3).

Since BX bisects angle ABC, then BX is also the altitude of triangle ABC. The area of a triangle is equal to 1/2baseheight. Since the base of triangle BXA is BX and the height of triangle BXA is BX, then the area of triangle BXA is 1/2BX^2 = 1/2(6sqrt(3))^2 = 18*3 = 54.

Therefore, the area of triangle BXA is 54.

I'm gonna round some numbers drastically, so  

the products are gonna be appriximations, but  

this is just to give you an idea of my approach.  

use tan(30^o) to find BC = 6.9  

area ABC = (1/2)(12)(6.9) = 41.6  

use tan(30^o) to find CX = 4  

area BXC = (1/2)(6.9)(4) = 13.7  

area BXA = area ABC – area BXC  

area BXA = 27.9  

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Hi, I hope this sketch helps:

Steps:

Ignore anything in red for now, we are going to find it.

But first, we need to identify what we really want. 
We want the area of Triangle BXA. To find the area of a triangle, it must have a base and a height.

The height is BC (The height is defined to be a vertical line connected the top most point of a triangle to the lower most point).

And the base (I.e. the side where the height is perpendicular on it) is AX.

Thus, the desired area is just: \(\dfrac{1}{2}*AX*BC\).

Now, we need to find both of BC and AX.

Firstly, for triangle ABC: we have AC = 12 and angle B is 60. 
So using trigonometry: \(\tan(60)=\dfrac{12}{BC} \implies BC=4\sqrt{3}\)

(Note: If trigonometry is not allowed, then you have to remember the ratios in the 30-60-90 triangle).

So we have found BC.

Next, consider triangle BXC:

Again, using trigonometry to find XC:

\(tan(30)=\dfrac{XC}{4\sqrt{3}}\implies XC=4\)

But we want AX not XC. Thus,  \(AX=12-4=8\)

Thus the area of the triangle is: \(\dfrac{1}{2}*8*4\sqrt{3}=16\sqrt{3}\)

I hope this helps!