Let [tex]\alpha[/tex] be the angle of an arc, [tex]\alpha =120^{\circ},[/tex] and r be the radius of the circle.
Then by the cosine rule,
[tex]\text{chord}^2=r^2+r^2-2\cdot r\cdot r\cdot \cos \alpha,\\ \\(8\sqrt{3})^2=2r^2-2r^2\cdot \cos 120^{\circ}=2r^2-2r^2\cdot \left(-\dfrac{1}{2}\right)=3r^2,\\ \\192=3r^2,\\ \\r^2=64,\\ \\r=8\ un.[/tex]
1. Find the area of the sector. Since [tex]\alpha =120^{\circ},[/tex] then
[tex]A_{sector}=\dfrac{\pi r^2}{3}=\dfrac{64\pi}{3}\ un^2.[/tex]
2. Find the area of the triangle formed by two radii and chord:
[tex]A_{triangle}=\dfrac{1}{2}\cdot r\cdot r\cdot \sin \alpha=\dfrac{64}{2}\cdot \dfrac{\sqrt{3}}{2}=16\sqrt{3}\ un^2.[/tex]
3. The area of the segment is
[tex]A_{segment}=A_{sector}-A_{triangle}=\dfrac{64\pi}{3}-16\sqrt{3}=\dfrac{64\pi-48\sqrt{3}}{3}\ un^2.[/tex]
Answer: [tex]\dfrac{64\pi-48\sqrt{3}}{3}\ un^2.[/tex]
