The angle $\angle ACB$ is the complement of $\angle ABC$, so $\sin \angle ACB = \sin (90^\circ - \angle ABC) = \boxed{\frac{3}{5}}$. We swapped the sine function because $\sin (90^\circ - x) = \sin (90^\circ + x) = \cos x$, and $\cos x = \sin (90^\circ - x)$.
