Applications of trigonometric functions

Solve the missing parts of the triangle satisfying

a=125
b=150
alpha=55 degrees
There are two solutions

1 answer

a/sin alpha = b/sin beta

125/sin 55 = 150/sin beta

sin beta = .819/125 * 150 = .9828

beta = 79.25 deg or 100.75 deg

gamma = 180 - (alpha+beta)
= 45.75 or 24.25
sin gamma = .716 or .411

c / sin gamma = 152.6

c = 152.6*.716 or 152.6*.411
= 109.26 or 62.72
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