In triangle ABC, #a=12.2, b=14.0#, #A=43°#, how do you find #B#?

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Answer 1 · 2018-05-19T21:45:15

#B = arcsin(14.0/12.2 sin 43^circ) #

#B approx 51.5^circ or B approx 128.5^circ #

# a/sin A = b/sin B#

# sin B = b/a sin A#

#sin B = 14.0/12.2 sin 43^circ #

#B = arcsin(14.0/12.2 sin 43^circ) #

#B approx 51.5^circ or B approx 180^circ - 51.5^circ = 128.5^circ #