+436 Find the area of triangle $ABC$ if $AB = 6,$ $BC = 8,$ and $\angle ACB = 30^\circ$.
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+129771 B
6 8
A C 30
We have a SSA situation......there may be 1 triangle formed, 2 triangles formed or no triangles formed
Law of Sines
sin A / 8 = sin 30 / 6
sin A / 8 = (1/2) / 6
sinA = (8/6) (1/2) = 2/3
arcsin (2/3) ≈ 41.81° or 180 - 41.81 = 138.19
So angle B = 180 - 30 - 41.81 = 108.19°
or angle B = 180 - 30 - 138.19 = 11.81°
If B = 108.19° area of ABC = (1/2) (8)(6)sin (108.19°) ≈ 22.8
If B = 11.81° area of ABC = (1/2) (8)(6) sin ( 11.81°) ≈ 4.9
