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). $$\angle C = 53.3$$ °, c = 127 cm, b = 147 cm
A) a = 135 cm, $\angle A = 58.6$ ° , $\angle B = 87.3$ °
B) a = 135 cm, $\angle A = 58.6$ °, $\angle B = 68.1$ °
C) a = 135 cm, $\angle A = 39.4$ °, $\angle B = 87.3$ °
D) a = 127 cm, $\angle A = 58.6$ °, $\angle B = 68.1$ °
E) a = 127 cm, $\angle A = 39.4$ °, $\angle B = 87.3$ °

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). $$\angle C = 53.3$$ °, c = 127 cm, b = 147 cm 
A) a = 135 cm, $\angle A = 58.6$ ° , $\angle B = 87.3$ ° 
B) a = 135 cm, $\angle A = 58.6$ °, $\angle B = 68.1$ ° 
C) a = 135 cm, $\angle A = 39.4$ °, $\angle B = 87.3$ ° 
D) a = 127 cm, $\angle A = 58.6$ °, $\angle B = 68.1$ ° 
E) a = 127 cm, $\angle A = 39.4$ °, $\angle B = 87.3$ °

Quizplus · Accepted Answer

Using the Law of Cosines, we have: 
$$a^2= b^2+c^2-2bc\cos{C}$$
$$a^2=147^2+127^2-2\cdot 147\cdot 127\cdot \cos(53.3) $$
$$a\approx 135$$

Now, using the Law of Sines, we have: 
$$\frac{a}{\sin{A}}=\frac{b}{\sin{B}} \Rightarrow \sin{B}=\frac{b\sin{A}}{a}$$
$$\cos{B}=\sqrt{1-\sin^2{B}}$$
$$\cos{B}=\frac{c^2+a^2-b^2}{2ac} \Rightarrow \sin{B}=\sqrt{1-\cos^2{B}}$$
$$\sin{B}=\frac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}}{2ac}$$

$$\sin{A}=\frac{a\sin{B}}{b} \Rightarrow \cos{A}=\sqrt{1-\sin^2{A}}$$
$$\cos{A}=\frac{b^2+a^2-c^2}{2ab} \Rightarrow \sin{A}=\sqrt{1-\cos^2{A}}$$

Using the values for a, b, c, and C that we just found, we get:
$$\sin{B} \approx 0.880$$
$$\cos{B} \approx 0.475$$
$$\sin{A} \approx 0.555$$
$$\cos{A} \approx 0.832$$

Therefore, we have: 
$$\angle A \approx 58.6 \text{°}$$
$$\angle B \approx 68.1 \text{°}$$

The sum of angles in a triangle is always 180°, so we can find the measure of angle C as follows: 
$$\angle C = 180 - \angle A - \angle B$$
$$\angle C \approx 53.3 \text{°}$$

Therefore, the answer is B.

Solve the Triangle Using the Labels as Shown in the Illustration