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). $\angle A = 19.6$ °, $\angle B = 37.1$ °, c = 41.7 cm
A) a = 30.1 cm, b = 16.7 cm, $\angle C = 123.3$ °
B) a = 16.7 cm, b = 30.1 cm, $\angle C = 123.3$ °
C) a = 30.1 cm, b = 16.7 cm, $\angle C = 19.6$ °
D) a = 16.7 cm, b = 30.1 cm, $\angle C = 19.6$ °
E) a = 16.7 cm, b = 30.1 cm, $\angle C = 37.1$ °

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). $\angle A = 19.6$ °, $\angle B = 37.1$ °, c = 41.7 cm 
A) a = 30.1 cm, b = 16.7 cm, $\angle C = 123.3$ ° 
B) a = 16.7 cm, b = 30.1 cm, $\angle C = 123.3$ ° 
C) a = 30.1 cm, b = 16.7 cm, $\angle C = 19.6$ ° 
D) a = 16.7 cm, b = 30.1 cm, $\angle C = 19.6$ ° 
E) a = 16.7 cm, b = 30.1 cm, $\angle C = 37.1$ °

Quizplus · Accepted Answer

Using the law of sines, we have:
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$
Substituting the given values, we have:
$$\frac{a}{\sin 19.6}=\frac{b}{\sin 37.1}=\frac{41.7}{\sin C}$$
Solving for $a$ and $b$ gives:
$$a=\sin 19.6 \cdot \frac{41.7}{\sin C}\approx 16.7	ext{ cm}$$
$$b=\sin 37.1 \cdot \frac{41.7}{\sin C}\approx 30.1	ext{ cm}$$
Now, using the fact that the angles of a triangle add up to $180^\circ$, we can solve for $\angle C$:
$$\angle C=180^\circ-\angle A-\angle B\approx 123.3^\circ$$
Therefore, the best choice is B, which matches these values.

Solve the Triangle Using the Labels as Shown in the Illustration