To find the area of triangle ABC with the given sides \( a = 57 \) m, \( b = 65 \) m, and included angle \( \angle C = 86^\circ \), we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2}ab \sin(C) \]
Substituting the values into the formula:
\[ \text{Area} = \frac{1}{2} \times 57 \times 65 \times \sin(86^\circ) \]
First, we calculate \( \sin(86^\circ) \). The sine of 86 degrees is approximately 0.998.
Now substituting this value:
\[ \text{Area} = \frac{1}{2} \times 57 \times 65 \times 0.998 \]
Calculating the product of \( 57 \) and \( 65 \):
\[ 57 \times 65 = 3705 \]
Now plug this into the area formula:
\[ \text{Area} \approx \frac{1}{2} \times 3705 \times 0.998 \] \[ \text{Area} \approx 1852.5 \times 0.998 \] \[ \text{Area} \approx 1850.563 \]
Now, rounding to the nearest square meter:
\[ \text{Area} \approx 1851 , \text{m}^2 \]
Thus, the area of triangle \( ABC \) to the nearest square meter is 1851 m².