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

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Answer 1 · 2016-12-18T13:53:20

#B = 51.5^@ (3s.f.)#

use the sine rule:

#sin A/a= sin B/b#

#sin A = sin 43^@#

#a = 12.2#

#b = 14.0#

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

multiply by 14:

#sin B = sin 43^@ * 14/12.2#

inverse sin:

#B = sin^-1 (sin 43^@ *14/12.2)#

#B = 51.5^@ (3s.f.)#