Let's assume that some triangle is described by its two sides and an angle (which is not between the given two sides however).
Basically for a triangle above only characteristics $c,\ b,\ C$ are known. Geometrically (by using visual imagination) it is clear to me that the resulting triangle is not always unique. Specifically for $C<90 \deg$ it is possible to construct both acute and obtuse triangles. Further complications include that apparently in some cases no triangle can be constructed at all.
What could be an algebraic proof of this non-uniqueness? I guess it would be an unpleasant procedure to try to explicitly derive $A$ value for example but maybe it's easier to show that this characteristic could have multiple solutions?
Start with point $A$ and place it in the origin of your coordinate system. Put point $C$ at distance $b$ from $A$. Describe the ray $\vec{CB}$ using $\angle C$. Now you have to intersect a circle of radius $c$ around point $A$ with that ray. Which is a quadratic equation that will in general have two solutions. Algebraically you can see this as two possible values for $a$.
$$ A = \begin{pmatrix}0\\0\end{pmatrix} \\ C = \begin{pmatrix}b\\0\end{pmatrix} \\ \vec{CB} = \left\{\begin{pmatrix}b\\0\end{pmatrix}+ \lambda\begin{pmatrix}-\cos\angle C\\\sin\angle C\end{pmatrix} \;\middle\vert\; \lambda\in\mathbb R\right\} \\ B = \begin{pmatrix}b\\0\end{pmatrix}+ a\begin{pmatrix}-\cos\angle C\\\sin\angle C\end{pmatrix} \\ \lVert A-B\rVert^2 = \lVert B\rVert^2 = (b-a\cos\angle C)^2+(a\sin\angle C)^2 = c^2 \\ a^2-2a\cos\angle C+b^2-c^2 = 0 \\ a_{1,2} = \cos\angle C\pm\sqrt{\cos^2\angle C-b^2+c^2} $$