Find the area of the triangle $A B C$.
\[
A=48.2^{\circ} \quad B=36.3^{\circ} \quad C=28.9 \mathrm{~m}
\]
What is the area of the triangle?
$\mathrm{m}^{2}$ (Round to the nearest tenth as needed.)

Step 1 ：We are given that the angles of the triangle are \(A = 48.2^\circ\), \(B = 36.3^\circ\), and the length of side \(C = 28.9 m\).

Step 2 ：We know that the sum of the angles in a triangle is 180 degrees. So we can calculate the third angle \(C\) as \(180 - A - B = 95.5^\circ\).

Step 3 ：We can use the Law of Sines to find the lengths of the other two sides. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. So, we can calculate the lengths of sides \(A\) and \(B\) as follows: \(A = C \cdot \frac{\sin(A)}{\sin(C)} = 21.64 m\) and \(B = C \cdot \frac{\sin(B)}{\sin(C)} = 17.19 m\).

Step 4 ：Once we have the lengths of the other two sides, we can use the formula for the area of a triangle to find the area. The area of a triangle can be calculated using the formula \(\frac{1}{2}ab\sin(C)\). Substituting the values we get, the area is approximately 185.15 square meters.

Step 5 ：Rounding to the nearest tenth, the area of the triangle is approximately \(\boxed{185.2}\) square meters.