Use the Law of Sines to solve (if possible) the triangle. Round your answers to two decimal places. $A = 76 ^ { \circ } , a = 102 , b = 63$
A) $ B \approx - 67 ^ { \circ } , C \approx 37 ^ { \circ } , c \approx 96.9$
B) $B \approx 73.2 ^ { \circ } , C \approx 30.8 ^ { \circ } , c \approx 96.9$
C) $B \approx 36.82 ^ { \circ } , C \approx 67.18 ^ { \circ } , c \approx 96.9$
D) $B \approx 67.18 ^ { \circ } , C \approx 36.82 ^ { \circ } , c \approx 96.9$
E) No Solution

Question

Use the Law of Sines to solve (if possible) the triangle. Round your answers to two decimal places. $A = 76 ^ { \circ } , a = 102 , b = 63$ 
A) $ B \approx - 67 ^ { \circ } , C \approx 37 ^ { \circ } , c \approx 96.9$
B) $B \approx 73.2 ^ { \circ } , C \approx 30.8 ^ { \circ } , c \approx 96.9$
C) $B \approx 36.82 ^ { \circ } , C \approx 67.18 ^ { \circ } , c \approx 96.9$
D) $B \approx 67.18 ^ { \circ } , C \approx 36.82 ^ { \circ } , c \approx 96.9$
E) No Solution

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:

$$\frac{102}{\sin 76 ^ {\circ}} = \frac{63}{\sin B} = \frac{c}{\sin C}$$

Solving for $\sin B$:

$$\sin B = \frac{63}{102} \sin 76 ^ {\circ} \approx 0.447$$

Using the inverse sine function, we get:

$$B \approx 27.38 ^ {\circ} 	ext{ or } 152.62 ^ {\circ}$$

Since $B$ is an angle in a triangle, it must be acute. Therefore, we have:

$$B \approx 27.38 ^ {\circ}$$

Similarly, solving for $\sin C$:

$$\sin C = \frac{63}{102} \frac{\sin 76^{\circ}}{\sin 27.38^{\circ}} \approx 0.768$$

Using the inverse sine function, we get:

$$C \approx 49.94^{\circ}$$

Finally, we can use the fact that the sum of the angles in a triangle is $180^{\circ}$ to find $A$:

$$A = 180^{\circ} - B - C \approx 103.68^{\circ}$$

We can check our work by verifying that the ratio of side lengths is consistent:

$$\frac{a}{\sin A} \approx \frac{102}{\sin 103.68^{\circ}} \approx 96.9 	ext{ (rounded)}$$

$$\frac{b}{\sin B} \approx \frac{63}{\sin 27.38^{\circ}} \approx 96.9 	ext{ (rounded)}$$

$$\frac{c}{\sin C} \approx \frac{96.9}{\sin 49.94^{\circ}} \approx 96.9 	ext{ (rounded)}$$

Therefore, the answer is (C).

Use the Law of Sines to Solve (If Possible) the Triangle