Well, I observe the triangle with side-lengths $a, b, c$ and angle $\alpha$, the opposite angle to $a$. The implicit relationship between them should be The Law of Cosines : $a^2-b^2-c^2+2bc\cos \alpha=0$
Then if I write $\alpha$ as smooth function of $a, b, c$, namely $\alpha=\arccos \frac{c^2+b^2-a^2}{2bc} = f(a,b,c)$, under what circumstances I can do that? This problem should be related to the Implicit Function Theorem. Of course, we must have $b,c\neq0$. Does that mean that $\alpha$ can be written as a smooth function of $a, b, c$ everywhere except in the neighbourhood of $(k,0,0)$, when we look from the point of IFT, or under another circumstances?
Put $F(a,b,c,\alpha) = a^2-b^2-c^2+2bc\cos\alpha$. Then $\alpha$ is defined implicitly as a function of $a,b,c$ by the equation $F(a,b,c,\alpha) = 0$. Examining carefully the statement of the Implicit Function Theorem, given $(a_0,b_0,c_0,\alpha_0)$ which satisfy the Law of Cosines, we may write $\alpha = f(a,b,c)$ (in such a way that $F(a,b,c, f(a,b,c)) = 0$) in a neighborhood of $(a_0,b_0,c_0)$ when $F_\alpha(a_0,b_0,c_0,\alpha_0)\neq 0$.
Since $F_\alpha = -2bc\sin\alpha$, we need $b,c\neq 0$ and $\alpha\neq 0$. In other words, we need $b,c>0$ and the triangle to not be degenerating (implying $a>0$ as well). So we are guaranteed invertibility so long as we can draw a valid triangle with positive side lengths $a,b,c$.