Question

If a = 57 m, b = 65 m, and angle C = 86 degrees, then what is the area of triangle ABC to the nearest square meter?

Answer 1

To find the area of triangle ABC with sides \( a \), \( b \), and included angle \( C \), we can use the formula:

\[
\text{Area} = \frac{1}{2}ab \sin(C)
\]

Given:
- \( a = 57 \, \text{m} \)
- \( b = 65 \, \text{m} \)
- \( C = 86^\circ \)

First, we need to calculate \( \sin(86^\circ) \). Using a calculator, we find:

\[
\sin(86^\circ) \approx 0.998
\]

Now, plug in the values into the area formula:

\[
\text{Area} = \frac{1}{2} \times 57 \, \text{m} \times 65 \, \text{m} \times \sin(86^\circ)
\]

Calculating the area:

\[
\text{Area} = \frac{1}{2} \times 57 \times 65 \times 0.998
\]

Calculating \( 57 \times 65 \):

\[
57 \times 65 = 3705
\]

Now calculate \( \frac{1}{2} \times 3705 \):

\[
\frac{1}{2} \times 3705 = 1852.5
\]

Now multiply by \( 0.998 \):

\[
1852.5 \times 0.998 \approx 1850.225
\]

Rounding this to the nearest square meter gives:

\[
\text{Area} \approx 1850 \, \text{m}^2
\]

Thus, the area of triangle ABC is approximately **1850 square meters**.