Let $O$ be the center of $(KMN)$ and $M'$ be the midpoint of $AM$. Let the line through $O$ parallel to $AB$ meet $CA$ and $CB$ at $A'$ and $B'$. Notice $M'O$ is the mid-line of $AMK$ so $M'O\parallel AK$. As $AM'\parallel A'O$ we have that $AM'OA'$ is a parallelogram. Now we are left to resolve configuration issues. Let $A'B'C$ have side length $x$. It is not hard to show that either $B'N=1$ or $B'N=x-2$. In the second case, $L$ would be the foot of $M$ onto $AC$ which does not lie outside $ABC$. So $B'N=1$ making $ONK$ equilateral so $\angle KMN=\boxed{30^{\circ}}$.

This seems easier than a few of the earlier ones. Consider $K \in BC$ such that $TK \perp AC$. Note that $TC=\frac{CK}{\cos 60}=2=MA$ $ \implies TM \parallel AC$ and $\angle KTM=90=\angle KNM$ $\implies KTNM$ is cyclic $\implies KMN=\angle CTK=30$