+721 Find the area of triangle $ABC$ if $AB = 6,$ $BC = 8,$ and $\angle ACB = 30^\circ$.
+129771 A
6
B 8 C 30
AB^2 = BC^2 + AC^2 - 2 (BC * AC) cos (30°)
36 = 64 + AC^2 - 2( 8 * AC) *(sqrt (3)/2)
-28 = AC^2 - 8sqrt (3)AC
AC^2 - 8sqrt (3) AC + 28 =
AC^2 - 8sqrt (3) AC = -28
AC^2 - 8sqrt (3) AC + 48 = -28 + 48
(AC - 4sqrt (3) )^2 = 20 take the positive root
AC - 4sqrt (3) = sqrt (20)
AC = sqrt (30) + 4sqrt 3 = sqrt (20) + sqrt (48)
[ABC] = (1/2) (BC)(AC) sin (ACB) =
(1/2) (8) ( sqrt (20) + sqrt (48) (1/2)
2 [ sqrt (20) + sqrt (48) ] = 2 [ 2sqrt (5) + 4sqrt (3) ] = 4sqrt (5) + 8sqrt (3) ≈ 22.8
