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). $\angle B = 40.8$ °, $\angle C = 61.3$ °, c = 22.7 km
A) a = 25.3 km, b = 16.9 km, $\angle A = 77.9$ °
B) a = 25.3 km, b = 16.9 km, $\angle A = 40.8$ °
C) a = 16.9 km, b = 25.3 km, $\angle A = 61.3$ °
D) a = 16.9 km, b = 25.3 km, $\angle A = 77.9$ °
E) a = 22.7 km, b = 16.9 km, $\angle A = 77.9$ °

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). $\angle B = 40.8$ °, $\angle C = 61.3$ °, c = 22.7 km 
A) a = 25.3 km, b = 16.9 km, $\angle A = 77.9$ ° 
B) a = 25.3 km, b = 16.9 km, $\angle A = 40.8$ ° 
C) a = 16.9 km, b = 25.3 km, $\angle A = 61.3$ ° 
D) a = 16.9 km, b = 25.3 km, $\angle A = 77.9$ ° 
E) a = 22.7 km, b = 16.9 km, $\angle A = 77.9$ °

Quizplus · Accepted Answer

We can use the Law of Sines to solve this triangle:
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$$
Plugging in the given values:
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin 40.8}=\dfrac{22.7}{\sin 61.3}$$
Solving for $b$ and $\sin A$:
$$b=\dfrac{\sin 40.8 \cdot 22.7}{\sin 61.3}=16.9 \text{ km}$$
$$\sin A=\dfrac{a\sin 77.9}{22.7} \implies a= \dfrac{22.7 \cdot \sin A}{\sin 77.9} = 25.3 \text{ km}$$
Finally, using the fact that the angles of a triangle sum to 180 degrees:
$$\angle A = 180 - \angle B - \angle C = 77.9 \text{ degrees}$$
Therefore, the correct answer is choice A.

Solve the Triangle Using the Labels as Shown in the Illustration