Get all dihedral angles of a tetrahedron given three angles and three edges Suppose you have a general tetrahedron, you are given the three angles that define a vertex of that tetrahedron. You also know the length of the edges that converge on this same vertex.
Given that, calculate all dihedral angles.
From This question and its accepted answer I can get three of the six dihedral angles. I am trying to figure out the rest (the ones at the "base").
I tried adapting the answer for that question, extending the definition of  $\vec v_1$, $\vec v_2$, $\vec v_3$, so that now they are not unit vectors, but their magnitude is the length of the edges.
Then we could similarly define a unit normal vector for the "base" as:
$$\vec n_{12}=\dfrac{\vec v_1 \times \vec v_2}{|\vec v_1 \times \vec v_2|}  \quad\text{and} \quad \vec n_{b}=\dfrac{\vec v_1 \times \vec v_2 + \vec v_2 \times \vec v_3 + \vec v_3 \times \vec v_1}{|\vec v_1 \times \vec v_2 + \vec v_2 \times \vec v_3 + \vec v_3 \times \vec v_1|}$$
so then, similar to that answer:
$$\text{cos}(θ_{ab})=-\vec n_{12}\cdot \vec n_{b}$$
But this hasn´t taken me very far. I feel like I might be very close but my vector arithmetics is too rusty. Or it is just simply wrong.
EDIT: Already solved this, thanks to your answers. However I am still trying to solve using my original approach. After some vector operations I get to this:
$\text{cos}(θ_{ab})=- \dfrac{a^2b^2\text{sin}^2\phi_{ab}+b^2ac(\text{cos}\phi_{ab}\text{cos}\phi_{bc}-\text{cos}\phi_{ac})+a^2bc(\text{cos}\phi_{ac}\text{cos}\phi_{ab}-\text{cos}\phi_{bc})}{ab\text{sin}\phi_{ab}(|\vec v_1 \times \vec v_2 + \vec v_2 \times \vec v_3 + \vec v_3 \times \vec v_1|)}$
Where $a,b,c$ are the lengths of the edges along $\vec v_1,\vec v_2,\vec v_3$ (also their magnitudes) and $\phi_{ab}$ is the angle between edges $a$ and $b$ (and so on)
So it is looking quite well, because it is mostly in terms of the angles and edge lengths, but still need to simplify those vectors. Any ideas?

I've used the Cartesian coordinate frame to find the coordinates of all the vertices of the tetrahedron.
Place the given vertex at the origin $(0, 0, 0)$.  Then the three edges meeting at the vertex are represented by the vectors $\vec{A}$ , $\vec{B}$ and $\vec{C}$.
We can take vector $\vec{A}$ to be along the $x$ axis, so we'll take point $A$ to be
$A = (a, 0, 0) $
where $a$ is the given length of this edge.
Next, we have point $B$, such that the angle between $\vec{B}$ and $\vec{A}$ is $\theta_{12}$, from this it follows that
$B = (b \cos \theta_{12}, b \sin \theta_{12} , 0 )$
Finally, we want to find the point $C$ along the third edge, so let
$C = (x, y, z)$
From the length requirement we know that $x^2 + y^2 + z^2 = c^2$
And from the angle constraints, we know that
$\vec{A} \cdot \vec{C} = a c \cos \theta_{13} = a x $
Hence $x = c \cos \theta_{13}$
Next, we know that
$\vec{B} \cdot \vec{C} = b c \cos \theta_{23} = b x \cos \theta_{12} + b y \sin \theta_{12} $
From this, it follows that,
$ y = c \dfrac{ \cos \theta_{23} - \cos \theta_{13} \cos \theta_{12} }{ \sin \theta_{12} }$
Using the length equation we can directly solve for $z$.
Now point $C$ is known.  So all the vertices of the tetrahedron have known coordinates.  One can now obtain the normals to the faces of the tetrahedron, then compute the dihedral angles in a direct way.
