How to calculate triangle side length and angle using sine method?

Question

The triangle dimensions are as follows:

For a=137mm, b=87mm, and C=44°, calculate the length of c. Then, calculate angle B.

I am unfamiliar with the sine method and have been thrown into a course requiring I use it. Any explanation on the answer and how to get the answer would be greatly appreciated.

Answer 1

To use the Law of Sines two angles must be known.

This problem presents only one.

Law of Cosines must be used first.

c² = a² + b² - 2ab cos C

c² = 137² + 87² - 2(137)(87)(0.6946)

c² = 9780.1252

c = 98.8945

Law of Sines

(a / sin A) = (b / sin B) = (c / sin C)

c / sin C = a / sin A

98.8945 / 0.6946 = 137 / sin A, solve for A

c / sin C = b / sin B

98.8945 / 0.6946 = 87 / sin B, solve for B

Answer 2

c^2 = a^2 + b^2 -2acCosC

c^2 = 137^2 + 87^2 -2(137)(87)Cos44

c = sqr[137^2 +87^2 -2(137)(87)Cos44= sqr(26338-17147.622)=95.8697

then calculate Sin44 = Cos46= .69465837

sin(x) = cos(90-x)

when you have calculated c, and SinC, then use the Law of Sines to solve for angles A and B

ratio of sides to opposite angles is a constant

c/Sin44 = b/SinB = a/SinA

c/Sin46 = 87/SinB = 137/SinA

95.8697/.694658 = 87/sinB

sinB =87(.69465837)/95.8697 = .63038

B=Sin-1(.6304) = about 39.07888 degrees

A should be 180-44-39.08 = 136-39.08= 96.92

c/Sin46 = 137/SinA

SinA = 137(Sin46)/c =137(.694658)/95.8697=.99268

A = Sin^-1(.9927)=about 83.06 degrees

there has to be an error in the above calculations, somewhere, as 83.06 doesn't = 96.92.

find the error and you score 2 points

Answer 3

I think you would need to use the law of cosines to solve. Law of sines is helpful when you have both a side and corresponding angle.

law of cosines is the formula:

c² = a² + b² - 2ab*cos(C)

c² = 137² + 87² - 2*87*137*cos(44)

c = 95.87

now that you have side c and angle C. You can use law of sines to solve for angle B

law of sines formula

sinB/b=sinC/c

sinB = b*sinC/c = 87*sin(44)/95.87 = 0.63

B = 39.08