+713 Find the area of triangle $ABC$ if $AB = 6,$ $BC = 8,$ and $\angle ACB = 30^\circ$
+129657 B
6 8
A C 30
Law pf Cosines
AB^2 = AC^2 + BC^2 - 2(AC * BC) cos (30)
6^2 = AC^2 + 8^2 - 2 (AC * 8C) sqrt (3)/2
36 - 64 = AC^2 - 8sqrt (3) AC
-28 = AC^2 - 8sqrt (3) AC
AC^2 - 8sqrt 3 AC + 28 = 0 solving for AC
AC ≈ 2.46 or AC ≈ 11.4
So..because pf the SSA situation.....two possible areas result
[ ABC] = (1/2) (2.46) (8) (1/2) ≈ 4.92
[ABC ] = (1/2) (11.4) (8)(1/2) ≈ 22.8
