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). $\angle B = 108.0$ °, c = 10.2 km, b = 31.0 km
A) a = 10.2 km, $\angle A = 18.2$ °, $\angle C = 53.8$ °
B) a = 31.0 km, $\angle A = 53.8$ °, $\angle C = 18.2$ °
C) a = 26.3 km, $\angle A = 53.8$ °, $\angle C = 18.2$ °
D) a = 31.0 km, $\angle A = 18.2$ °, $\angle C = 53.8$ °
E) a = 26.3 km, $\angle A = 18.2$ °, $\angle C = 53.8$ °

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). $\angle B = 108.0$ °, c = 10.2 km, b = 31.0 km 
A) a = 10.2 km, $\angle A = 18.2$ °, $\angle C = 53.8$ ° 
B) a = 31.0 km, $\angle A = 53.8$ °, $\angle C = 18.2$ ° 
C) a = 26.3 km, $\angle A = 53.8$ °, $\angle C = 18.2$ ° 
D) a = 31.0 km, $\angle A = 18.2$ °, $\angle C = 53.8$ ° 
E) a = 26.3 km, $\angle A = 18.2$ °, $\angle C = 53.8$ °

Quizplus · Accepted Answer

Using the law of cosines, we have:
$a^2 = b^2 + c^2 - 2bc\cos{B}$
$a^2 = (31.0 	ext{ km})^2 + (10.2	ext{ km})^2 - 2(31.0	ext{ km})(10.2	ext{ km})\cos{108.0^\circ}$
$a^2 = 862.17	ext{ km}^2$
$a = \sqrt{862.17}	ext{ km} \approx 26.3	ext{ km}$
Now, using the law of sines, we have:
$\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}}$
$\frac{26.3	ext{ km}}{\sin{A}} = \frac{31.0	ext{ km}}{\sin{108.0^\circ}} = \frac{10.2	ext{ km}}{\sin{C}}$
$\sin{A} \approx 0.309$
$\sin{C} \approx 0.101$
Note that $\sin{53.8^\circ} \approx 0.8$ and $\sin{18.2^\circ} \approx 0.309$. Therefore, the solution is:
$a = 26.3	ext{ km}, \angle A \approx 18.2^\circ, \angle C \approx 53.8^\circ$

Solve the Triangle Using the Labels as Shown in the Illustration