How do you find the angle of A when B is 100 degrees, C is unknown, and it is not a right triangle?

Whether to use the Law of Sines or Cosines will really depend on what other data are given.

Based on what you have posted, there is no sufficient information as to what formula needs to be used.

Hope this helps.

You need minimum of three elements to solve any triangle.

The one you show ("sinC/A = sinA/C") is none that I know of. SinA/a = SinC/c is the Sine law but that won't work here, at least at first.. Cosine law will.
(AC)^2 = (AB)^2 + (BC)^2 - 2(AB)(BC)CosB
(AC)^2 = (3)^2 + (6)^2 - 2(3)(6)(-0.1736)
(AC)^2 = 9 + 36 + 6.2496
(AC)^2 = 51.2496
AC = 7.16
Now Sine law
Sin100/7.16 = SinA/6
0.9848/7.16 = SinA/6........cross multiply
7.16SinA = 5.909
SinA = 0.8252
A = 55.6 degrees......answer
Sin100/7.16 = SinC/3
7.16SinC = (0.9848)(3)
SinC = 2.9544/7.16
C = 24.4 degrees.........answer

Check
55.6 + 100 + 24.4 = 180
180 = 180