Dimension of configuration space of triangulated convex polyhedron 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?
 A: A convex polyhedron can be considered as a flat metric with conic singularities
on the sphere. This metric is completely determined (up to a constant factor)
by conformal structure of the sphere with $v$ marked points ($v$ is the number of vertices) and $v$ angles around these points (sums of angles of faces meeting at a vertex). Three points can be fixed and the rest
depend of $2$ real parameters each. The angles satisfy one relation (coming from Gauss-Bonnet). So the total number of parameters is
$$2(v-3)+v-1=3v-7.$$
For the tetrahedron, $v=4$ and we obtain $5$.
(Since your argument involves only angles, I suppose you counted tetrahedra up to scaling, as I did).
The fact that every flat metric with conic singularities and angles
$<2\pi$ at each singularity corresponds to a convex polytope embedded in $R^3$ is non-trivial, but I suppose it is true.
Remark: Your condition that it is "triangulated", if I understand it
correctly, is irrelevant: the surface of any convex polytope can be triangulated without adding new vertices.
Remark 2: this answer is less elementary but contains some extra information in comparison with Yoav Kallus's answer: it gives a global parametrization of the set of convex polytopes, not only the dimension count. It also works for hyperbolic or spherical polytopes (assuming that the angles are less than $2\pi$.
A: For simplicial polyhedra, one can simply add up the degrees of freedom from placing the vertices in space and substract 7 degrees of freedom for rotation, translation, and scaling to obtain $3v-7$ as in Alexandre Eremenko's answer.
