Solve the triangle using the labels as shown in illustration (round lengths of sides to three significant digits and angles to the nearest tenth of a degree). $\angle A = 53.0$ °, $\angle C = 48.1$ °, a = 26.5 m
A) b = 32.6 m, c = 26.5 m, $\angle B = 78.9$ °
B) b = 32.6 m, c = 24.7 m, $\angle B = 78.9$ °
C) b = 24.7 m, c = 32.6 m, $\angle B = 78.9$ °
D) b = 32.6 m, c = 24.7 m, $\angle B = 48.1$ °
E) b = 24.7 m, c = 32.6 m, $\angle B = 48.1$ °

Question

Solve the triangle using the labels as shown in illustration (round lengths of sides to three significant digits and angles to the nearest tenth of a degree). $\angle A = 53.0$ °, $\angle C = 48.1$ °, a = 26.5 m 
A) b = 32.6 m, c = 26.5 m, $\angle B = 78.9$ ° 
B) b = 32.6 m, c = 24.7 m, $\angle B = 78.9$ ° 
C) b = 24.7 m, c = 32.6 m, $\angle B = 78.9$ ° 
D) b = 32.6 m, c = 24.7 m, $\angle B = 48.1$ ° 
E) b = 24.7 m, c = 32.6 m, $\angle B = 48.1$ °

Quizplus · Accepted Answer

To solve the triangle, we can use the law of sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

Substituting the given values, we have:

$$\frac{26.5}{\sin 53.0} = \frac{b}{\sin B} = \frac{c}{\sin 48.1}$$

Solving for either $b$ or $c$, we get:

$$b = \frac{\sin 53.0}{\sin 78.9} \cdot 26.5 \approx 32.6 \text{ m}$$
$$c = \frac{\sin 48.1}{\sin 78.9} \cdot 26.5 \approx 24.7 \text{ m}$$

To find $\angle B$, we can use the law of cosines:

$$a^2 = b^2 + c^2 - 2bc \cos A$$

Substituting the given values, we have:

$$26.5^2 = (32.6)^2 + (24.7)^2 - 2(32.6)(24.7) \cos 53.0$$

Solving for $\cos 53.0$, we get:

$$\cos 53.0 = \frac{(32.6)^2 + (24.7)^2 - 26.5^2}{2(32.6)(24.7)} \approx 0.478$$

Taking the inverse cosine, we get:

$$\angle B \approx \cos^{-1} 0.478 \approx 78.9 \text{°}$$

Therefore, the best choice is (B), with $b \approx 32.6$ m, $c \approx 24.7$ m, and $\angle B \approx 78.9$°.

Solve the Triangle Using the Labels as Shown in Illustration ∠A=53.0\angle A = 53.0∠A=53.0

Solve the Triangle Using the Labels as Shown in Illustration $\angle A = 53.0$