A geometry problem by d w

Using the sine formula for calculating the area of a triangle:

$A=$ $\frac{1}{2}ab\ sinC°$

It becomes:

$0.5*4*4*sin\ 120°$

Which equals to

$4√3$

If we bisect the 120 degree angle, we will get two identical right triangles. The right triangles formed are 30 60 90's, a special right triangle with sides $\sqrt{3}x,x, 2x$ . And $4=2x$ , so $x=2$ . Now we take the area of one of the triangles, and double it, we will have the area: $\frac{2\cdot2\sqrt{3}}{2}\cdot2=4\sqrt{3}$ .