Use the law of Cosines to solve the given triangle.Round your answer to two decimal places.
C = 40°, a = $\frac { 2 } { 15 }$ , b = $\frac { 1 } { 5 }$
A)A $\approx$ 41.21°, B $\approx$ 98.79°, c $\approx$ 0.13
B)A $\approx$ 99.76°, B $\approx$ 98.76°, c $\approx$0.13
C)A $\approx$ 98.76°, B $\approx$ 41.24°, c $\approx$ 0.20
D)A $\approx$ 41.24°, B $\approx$ 98.76°, c $\approx$ 0.20
E)A $\approx$ 98.79°, B $\approx$ 41.21°, c $\approx$ 0.20

Question

Use the law of Cosines to solve the given triangle.Round your answer to two decimal places.  
C = 40°, a = $\frac { 2 } { 15 }$ , b = $\frac { 1 } { 5 }$  
A)A $\approx$ 41.21°, B $\approx$ 98.79°, c $\approx$ 0.13
B)A $\approx$ 99.76°, B $\approx$ 98.76°, c $\approx$0.13
C)A $\approx$ 98.76°, B $\approx$ 41.24°, c $\approx$ 0.20
D)A $\approx$ 41.24°, B $\approx$ 98.76°, c $\approx$ 0.20
E)A $\approx$ 98.79°, B $\approx$ 41.21°, c $\approx$ 0.20

Quizplus · Accepted Answer

The law of cosines is used to find the third side of the triangle, $c^2 = a^2 + b^2 - 2ab\cos(C)$. Substituting the given values, $c \approx 0.13$. Then, using the law of sines or further application of the law of cosines, angles A and B can be found, leading to the conclusion that A $\approx$ 41.21° and B $\approx$ 98.79°.

Use the Law of Cosines to Solve the Given Triangle 215\frac { 2 } { 15 }152​

Use the Law of Cosines to Solve the Given Triangle $\frac { 2 } { 15 }$