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Find the area of triangle ABC.

Given:
- Angle A = 44.8º
- Angle B = 39.3º
- Side c = 21.8 m

What is the area of the triangle? (Round to the nearest tenth as needed.)

Answer :

To find triangle area, we use the formula A = 1/2 * base * height. We first find the third angle of the triangle by subtracting the sum of the given angles from 180 degrees. Then, we use the Law of Sines to find the length of the base. Therefore, the area of triangle ABC is approximately 187.7 square meters.

To find the area of a triangle, we can use the formula A = 1/2 * base * height. In this case, the base can be any side of the triangle and the height is the perpendicular distance from the base to the opposite vertex. Since we are not given the base and height directly, we need to use the given angles and side lengths to find them.

First, let's find the third angle of the triangle by subtracting the sum of the given angles from 180 degrees. The third angle would be 180 - 44.8 - 39.3 = 96.9 degrees.

Next, we can use the Law of Sines to find the length of the base. The Law of Sines states that a/sin(A) = b/sin(B) = c/sin(C), where a, b, and c are the side lengths and A, B, and C are the angles opposite the corresponding sides. Let's choose side BC as the base, with side lengths c = 21.8 m and C = 96.9 degrees. Using the law of sines, we can solve for side a as follows:

a/sin(A) = c/sin(C)

a/sin(44.8) = 21.8/sin(96.9)

a = 21.8 * sin(44.8) / sin(96.9)

= 16.3 m

Now that we have the base and the height (which is the opposite side of the given angle), we can use the formula A = 1/2 * base * height:

A = 1/2 * 16.3 * 21.8

= 187.7

Therefore, the area of triangle ABC is approximately 187.7 square meters.

Learn more about the topic of Triangle Area here:

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