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

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

$B approx 51.5^circ or B approx 128.5^circ$

$a/sin A = b/sin B$

$sin B = b/a sin A$

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

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

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

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