In order to better understand, let's draw a figure and include a triangle created by 2 radius and the given chord:
Since the triangle is isosceles, we can calculate the radius dividing the triangle in two with the height relative to its base:
Using the sine relation of the 60° angle, we have:
[tex]\begin{gathered} \sin (60\degree)=\frac{4\sqrt[]{3}}{r} \\ \frac{\sqrt[]{3}}{2}=\frac{4\sqrt[]{3}}{r} \\ \frac{1}{2}=\frac{4}{r} \\ r=8 \end{gathered}[/tex]
Now, let's find the height and then calculate the triangle area:
[tex]\begin{gathered} \cos (60\degree)=\frac{h}{r} \\ \frac{1}{2}=\frac{h}{8} \\ h=4 \\ \\ A=\frac{b\cdot h}{2} \\ A=\frac{8\sqrt[]{3}\cdot4}{2}=16\sqrt[]{3} \end{gathered}[/tex]
Now, calculating the area of the segment, we have:
[tex]\begin{gathered} A=\frac{\theta}{360}\cdot\pi r^2-A_{triangle} \\ A=\frac{120}{360}\cdot\pi\cdot8^2-16\sqrt[]{3} \\ A=\frac{1}{3}\cdot64\pi-16\sqrt[]{3} \\ A=\frac{64\pi}{3}-16\sqrt[]{3} \end{gathered}[/tex]
