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need more explanation to this review/exercise so I can understand what I’m going to do in my test.

Need More Explanation To This Reviewexercise So I Can Understand What Im Going To Do In My Test class=

Sagot :

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.

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