How do you solve the triangle given #B=28^circ, C=104^circ, a=3 5/8#?

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Answer 1 · 2018-02-18T14:24:10

#a = 3(5/8), b = 2.29, c = 4.733#

#hatA = 48^@, hatB = 28^@, hatC = 104^@#

#hatB = 28, hatC = 104, a = 29/8#

To find #hatA, b, c#

#hatA = 180 - 28 - 104 = 48^@#

#a / sin A = b / sin B = c / sin C#

#b = (a * sin B) / sin A = ((29/8)*sin 28) / sin 48 = 2.29#

#c = (a * sin C) / sin A = ((29/8)*sin 104) / sin 48 = 4.733#