Maximum Angle ACD in Triangle

$D$ is the midpoint of $AB$, so $AD = 0.5$. To maximize $\angle ACD$, $CD \perp AB$ and $AC$ should be as small as possible. Since $AC > 1$ and integer-valued, minimum $AC = 2$. In right triangle $ADC$, $\sin(\angle ACD) = \frac{AD}{AC} = \frac{0.5}{2} = 0.25$. $\angle ACD = \arcsin(0.25) \approx 14.48^\circ \approx 15^\circ$.