Solve the right triangle. Round your answer to the appropriate number of significant digits. $b = 71.0 \mathrm { mi } , c = 77.0 \mathrm { mi }$
A) $a = 29.8 \mathrm { mi } , \angle A = 52.3 ^ { \circ } , \angle B = 37.7 ^ { \circ }$
B) $a = 29.8 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 112.8 ^ { \circ }$
C) $a = 29.8 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 67.2 ^ { \circ }$
D) $a = 77.0 \mathrm { mi } , \angle A = 67.2 ^ { \circ } , \angle B = 22.8 ^ { \circ }$
E) $a = 71.0 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 67.2 ^ { \circ }$

Question

Solve the right triangle. Round your answer to the appropriate number of significant digits. $b = 71.0 \mathrm { mi } , c = 77.0 \mathrm { mi }$ 
A) $a = 29.8 \mathrm { mi } , \angle A = 52.3 ^ { \circ } , \angle B = 37.7 ^ { \circ }$ 
B) $a = 29.8 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 112.8 ^ { \circ }$ 
C) $a = 29.8 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 67.2 ^ { \circ }$ 
D) $a = 77.0 \mathrm { mi } , \angle A = 67.2 ^ { \circ } , \angle B = 22.8 ^ { \circ }$ 
E) $a = 71.0 \mathrm { mi } , \angle A = 22.8 ^ { \circ } , \angle B = 67.2 ^ { \circ }$

Quizplus · Accepted Answer

Using the Pythagorean theorem, we have:

$a = \sqrt{c^2 - b^2} = \sqrt{(77.0 \mathrm{mi})^2 - (71.0 \mathrm{mi})^2} \approx 29.8 \mathrm{mi}$

Then, using the sine law:

$\frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \sin A = \frac{a \sin B}{b} = \frac{(29.8 \mathrm{mi}) \sin^{-1} \left(\frac{71.0 \mathrm{mi}}{77.0 \mathrm{mi}}\right)}{71.0 \mathrm{mi}} \approx 0.4046$

Taking the inverse sine of both sides, we get:

$A \approx 22.8^\circ$

Finally, using the fact that the angles of a triangle sum to $180^\circ$, we have:

$B = 180^\circ - A - C \approx 67.2^\circ$

So the correct answer is C.

Solve the Right Triangle b=71.0mi,c=77.0mib = 71.0 \mathrm { mi } , c = 77.0 \mathrm { mi }b=71.0mi,c=77.0mi

Solve the Right Triangle $b = 71.0 \mathrm { mi } , c = 77.0 \mathrm { mi }$