Solve the triangle using the labels as shown in the illustration (round lengths of sides to three significant digits and angles to the nearest tenth of a degree). $\angle B = 74.6$ °, $\angle A = 63.0$ °, b = 302 ft
A) a = 211 ft, c = 279 ft, $\angle C = 42.4$ °
B) a = 279 ft, c = 211 ft, $\angle C = 42.4$ °
C) a = 279 ft, c = 211 ft, $\angle C = 63.0$ °
D) a = 279 ft, c = 302 ft, $\angle C = 42.4$ °
E) a = 211 ft, c = 279 ft, $\angle C = 63.0$ °

Question

Solve the triangle using the labels as shown in the illustration (round lengths of sides to three significant digits and angles to the nearest tenth of a degree). $\angle B = 74.6$ °, $\angle A = 63.0$ °, b = 302 ft 
A) a = 211 ft, c = 279 ft, $\angle C = 42.4$ ° 
B) a = 279 ft, c = 211 ft, $\angle C = 42.4$ ° 
C) a = 279 ft, c = 211 ft, $\angle C = 63.0$ ° 
D) a = 279 ft, c = 302 ft, $\angle C = 42.4$ ° 
E) a = 211 ft, c = 279 ft, $\angle C = 63.0$ °

Quizplus · Accepted Answer

We can use the Law of Sines to solve for the remaining side and angle. 
$$\frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow a = \frac{b\sin A}{\sin B} = \frac{302\sin 63}{\sin 74.6} \approx 279\text{ ft}$$
$$\frac{c}{\sin C} = \frac{b}{\sin B} \Rightarrow c = \frac{b\sin C}{\sin B} = \frac{302\sin (180 - A - B)}{\sin B} \approx 211\text{ ft}$$
$$\angle C = 180 - \angle A - \angle B \approx 42.4^\circ$$
Therefore, the only option that matches the calculated values is choice B.

Solve the Triangle Using the Labels as Shown in the Illustration