Given a figure of triangle ABC, whose side BC is 12 and angle A and angle B are 60 degrees.
We have to find the area of a triangle.
The area of the triangle ABC is:
[tex]A=\frac{1}{2}\times CD\times AB[/tex]
Here, we'll use the trigonometric function sine to find the length of the side CD.
In triangle BDC,
[tex]\begin{gathered} \sin 60=\frac{CD}{BC} \\ \frac{\sqrt[]{3}}{2}=\frac{CD}{12} \\ CD=\frac{12\sqrt[]{3}}{2} \\ CD=6\sqrt[]{3} \end{gathered}[/tex]
Since two angles of the triangle are given 60 degrees. So, the triangle ABC is an equilateral triangle. Therefore, AB = BC = 12.
Now, the area will be:
[tex]\begin{gathered} A=\frac{1}{2}\times6\sqrt[]{3}\times12 \\ =36\sqrt[]{3} \end{gathered}[/tex]
Thus, option 3 is correct.
