Well, here's one way to do it at least...
For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB).
Let P=(4,0) be the projection of B onto the x-axis.
Let Q=(-3,0) be the projection of C onto the x-axis.
Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4).
Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2).
Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go.
||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC)
||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB)
Using the Law of Cosines...
||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A)
||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191)
||a||^2 = 86 - 36
||a||^2 = 50
||a|| = sqrt(50)
Now apply the Law of Sines to find the other two angles.
||b|| / sin(B) = ||a|| / sin(A)
sqrt(41) / sin(B) = sqrt(50) / .9080
(.9080)sqrt(41) / sqrt(50) = sin(B)
.8222 = sin(B)
asin(.8222) = B
55.305 = B
Two down, one to go...
||c|| / sin(C) = ||a|| / sin(A)
sqrt(45) / sin(C) = sqrt(50) / .9080
(.9080)sqrt(45) / sqrt(50) = sin(C)
.8614 = sin(C)
asin(.8614) = C
59.470 = C
So your three angles are:
A = 65.225
B = 55.305
C = 59.470
