How do you solve the #triangle ABC# given #C=84^circ, c=7, a=2#?

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Answer 1 · 2018-02-24T13:07:01

#a = 2, b = 6.92, c = 7#

#hatA = 16.5^@, hatB = 79.5^@, hatC = 84^@#

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Given : #hatC = 84^@, a = 2, c = 7#

Applying law of sines,

#sin A / a = sin C / c#

#sin A = (a sin C) / sin A = (2 * sin 84) //7 = 0.2841#

#hatA = sin ^-1 (0.2841) ~~ 16.5^@#

Sum of the three angles of a triangle equals #180^@#

Hence #hatB = 180 - 84 - 16.5 = 79.5^@#

#b = (c sin B) / sin C = (7 * sin 79.5) / sin 84 = 6.92#