Solve the right triangle.
-$$\mathrm { A } = 57 ^ { \circ } 4 ^ { \prime } , \mathrm { c } = 242 \mathrm {~m} , \mathrm { C } = 90 ^ { \circ }$$
Round side lengths to two decimal places, if necessary.
A) $\mathrm { B } = 33 ^ { \circ } 46 ^ { \prime } , \mathrm { a } = 203.31 \mathrm {~m} , \mathrm {~b} = 131.57 \mathrm {~m}$
B) $\mathrm { B } = 32 ^ { \circ } 4 ^ { \prime } , \mathrm { a } = 204.31 \mathrm {~m} , \mathrm {~b} = 124.57 \mathrm {~m}$
C) $\mathrm { B } = 32 ^ { \circ } 56 ^ { \prime } , \mathrm { a } = 203.31 \mathrm {~m} , \mathrm {~b} = 138.57 \mathrm {~m}$
D) $\mathrm { B } = 32 ^ { \circ } 56 ^ { \prime } , \mathrm { a } = 203.11 \mathrm {~m} , \mathrm {~b} = 131.57 \mathrm {~m}$

Question

Quizplus · Accepted Answer

Angle B is found by subtracting angle A from 90 degrees, giving B = 32° 56'. Using the sine and cosine functions with the hypotenuse c = 242 m, side a = 242 * sin(57° 4') = 203.11 m and side b = 242 * cos(57° 4') = 131.57 m.

Solve the Right Triangle A=57∘4′,c=242 m,C=90∘\mathrm { A } = 57 ^ { \circ } 4 ^ { \prime } , \mathrm { c } = 242 \mathrm {~m} , \mathrm { C } = 90 ^ { \circ }A=57∘4′,c=242 m,C=90∘

Solve the Right Triangle $\mathrm { A } = 57 ^ { \circ } 4 ^ { \prime } , \mathrm { c } = 242 \mathrm {~m} , \mathrm { C } = 90 ^ { \circ }$