The configuration space of all tetrahedra is $5$-dimensional, perhaps a non-obvious fact. There are $12$ face angles, but the sum of each of the four faces angles is $\pi$, reducing $12$ to $8$ degrees of freedom. There are, however, additional trigonometric relations that must be satisfied by the angles, as cited in the Wikipedia article. This reduces the dimension of the configuration space to $5$ dimensions.
Wikipedia image: product of sines of marked angles are equal.
My question is:
Is there a generalization to other triangulated convex polyhedra?
For example, consider the space of all convex polyhedra that are combinatorially equivalent to a regular octahedron. Here we have $24$ face angles ($4$ per vertex), but then $8$ triangle angles summing to $\pi$ reduces the $24$ to $16$ degrees of freedom. Presumably there are additional trigonometric relations that further reduce the dimension of the configuration space. My guess is to $10$ dimensions.
Perhaps it is better to think in terms of vertex coordinates rather than in terms of face angles?