How do you solve this triangle by using the law of sine where
a= 6 m
b= 4 m
theta= 60°

Actually, when you know 2 sides and an included angle, you use the Law of Cosines. (and we don't know if theta is an included angle).
Solving for side c
c^2 = a^2 + b^2 -2ab * cos(C)
c^2 = 36 + 16 - 2*6*4 * cos(60)
c^2 = 52 -48*.5
c^2 = 28
c = 5.2915

Using the Law of Sines
side c / sin(C) = side b / sin (B)
5.2915 / sin(60) = 4 / sin (B)
sin(B) = sin(60) * 4 / 5.2915
sin(B) = 0.86603 * 4 / 5.2915
sin(B) = 3.46412 / 5.2915
sin(B) = 0.6546571451  
Angle B = 40.894 Degrees

sin (A) / side a = sin (B) / side b
sin (A) = 6 * sin (40.894) / 4
sin (A) = 6 * 0.65466 / 4
sin (A) = .98199
angle A = 79.109 Degrees

angle C = 60 Degrees