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