Find the area of triangle ABC. Give your answer correct to 1 decimal place.

Side AC: 12 cm
Angle A: 72°
Angle B: 59°

A. 32.4 cm²
B. 37.6 cm²
C. 41.8 cm²
D. 47.0 cm²

Question

High School

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Answer :

ArloHugo ArloHugo · Answer 1 · 2024-06-21T12:37:42-04:00

Rounded to 1 decimal place, the area of triangle ABC is approximately[tex]$ 61.6 \, \text{cm}^2 $.[/tex]

Given:

- Side AC = 12 cm

- Angle A = 72° (included angle)

- Angle B = 59° (included angle)

We'll first find side AB using the law of sines. Then we'll use the formula for the area of a triangle.

Using the law of sines:

[tex]\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} \][/tex]

[tex]\[ \frac{\sin(72°)}{12} = \frac{\sin(59°)}{AB} \][/tex]

[tex]\[ AB = \frac{12 \times \sin(59°)}{\sin(72°)} \][/tex]

[tex]\[ AB \approx \frac{12 \times 0.857}{0.951} \][/tex]

[tex]\[ AB \approx \frac{10.28}{0.951} \][/tex]

[tex]\[ AB \approx 10.8 \, \text{cm} \][/tex]

Now, using the formula for the area of a triangle:

[tex]\[ \text{Area} = \frac{1}{2} \times 12 \times 10.8 \times \sin(72°) \][/tex]

[tex]\[ \text{Area} \approx \frac{1}{2} \times 12 \times 10.8 \times 0.951 \][/tex]

[tex]\[ \text{Area} \approx \frac{1}{2} \times 129.6 \times 0.951 \][/tex]

[tex]\[ \text{Area} \approx \frac{1}{2} \times 123.144 \][/tex]

[tex]\[ \text{Area} \approx 61.572 \, \text{cm}^2 \][/tex]

[tex]$ 61.6 \, \text{cm}^2 $.[/tex]