Question

Use the law of sines to solve the triangle B=31, C=104, a=3(5/8)m

Answer 1

SinA / a = SinB / b = SinC / c

if ∠C = 104 + ∠B = 31, therefore ∠A = 45  (A + B + C = 180)

Sin45 / 3 5/8 = Sin31 / b      (multiply both sides by b)
bSin45 / 3 5/8 = Sin31          (multiply both sides by 3 5/8)
bSin45 = 3 5/8Sin31            (divide by Sin45)
b = 3.625Sin31 / sin45
b = 2.64

Sin45 / 3 5/8 = Sin104 / c      (multiply by c)
cSin45 / 3 5/8 = Sin104        (multiply both sides by 3 5/8)
cSin45 = 3 5/8Sin104            (divide by Sin45)
c = 3.62Sin104 / Sin45
c = 4.97

A = 45
b = 2.64
c = 4.97