Question

find the area of the triangle abc.
a = 47.2° b = 38.7° c = 20.2 m

the area of the triangle is □m².
(round to the nearest tenth as needed.)

Explanation:

Step1: Find angle \( C \)

Since the sum of angles in a triangle is \( 180^{\circ} \), \( C=180^{\circ}-A - B \).
Substitute \( A = 47.2^{\circ} \) and \( B = 38.7^{\circ} \):
\( C=180^{\circ}-47.2^{\circ}-38.7^{\circ}=94.1^{\circ} \)

Step2: Use the sine rule to find sides \( a \) and \( b \)

By the sine rule \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \).
First, find \( a \): \( a=\frac{c\sin A}{\sin C}=\frac{20.2\times\sin47.2^{\circ}}{\sin94.1^{\circ}} \)
\(\sin47.2^{\circ}\approx0.734\), \(\sin94.1^{\circ}\approx0.997\)
\( a=\frac{20.2\times0.734}{0.997}\approx14.8\)

Find \( b \): \( b = \frac{c\sin B}{\sin C}=\frac{20.2\times\sin38.7^{\circ}}{\sin94.1^{\circ}} \)
\(\sin38.7^{\circ}\approx0.625\)
\( b=\frac{20.2\times0.625}{0.997}\approx12.7\)

Step3: Calculate the area using the formula \( S=\frac{1}{2}ab\sin C \)

Substitute \( a = 14.8\), \( b = 12.7\), \( C = 94.1^{\circ}\) (\(\sin94.1^{\circ}\approx0.997\))
\( S=\frac{1}{2}\times14.8\times12.7\times0.997 \)
\( S=\frac{1}{2}\times14.8\times12.7\times0.997\approx93.6\)

Answer:

\( 93.6 \)