The Dihedral Angles of a Tetrahedron in terms of its edge lengths I am interested in any references which discuss a general formula for the dihedral angles of a tetrahedron in terms of its six edge lengths. If there is a well known formula could someone please post it here.
Edit: The solution below works in the case of a Euclidean tetrahedron, which I am thankful for. Is anyone aware of other methods that extend to higher dimensions, i.e. like the Cayley-Menger method for computing volumes does? I should also mention that I am interested in the cosine of the dihedral angle, the sine is easy to find using the generalized sine law. 
 A: The dihedral angles of a tetrahedron are related to the areas of the faces and "pseudo-faces" of the tetrahedron in a strikingly-familiar way: 
A Law of Cosines for Tetrahedra
$$\begin{eqnarray*}
W^2 + X^2 - 2 W X \cos \angle BC &= H^2 =& Y^2 + Z^2 - 2 Y Z \cos \angle OA \\
W^2 + Y^2 - 2 W Y \cos \angle CA &= J^2 =& Z^2 + X^2 - 2 Z X \cos \angle OB \\
W^2 + Z^2 - 2 W Z \cos \angle AB &= K^2 =& X^2 + Y^2 - 2 X Y \cos \angle OC
\end{eqnarray*}$$
Here, $W$, $X$, $Y$, and $Z$ are the faces of tetrahedron $OABC$, such that faces $W$ and $X$ share edge $BC$; faces $Y$ and $Z$ share edge $OA$; etc. (Note the opposition of edges: $BC$ is opposite $OA$, etc.) Naturally, "$\angle UV$" indicates the dihedral angle along edge $UV$.
$H$, $J$, and $K$ are (what I call) "pseudo-faces" of the tetrahedron. They're related to projections of the tetrahedron in planes parallel to opposite edges: for example, $H$ is a (possibly non-convex) quadrilateral whose diagonals are the projections of $OA$ and $BC$ into the plane parallel to those edges. (When $H$ is convex, it's the shadow of the tetrahedron in that plane.)
To answer your question ...
Use Heron's formula to compute the areas $W$, $X$, $Y$, and $Z$ from the lengths of edges. Use the following to compute $H$, $J$, and $K$:
$$\begin{eqnarray*}
H^2 = \frac{1}{16}\left( 4 a^2 d^2 - \left(( b^2 + e^2 ) - ( c^2 + f^2 )\right)^2 \right) \\
J^2 = \frac{1}{16}\left( 4 b^2 e^2 - \left(( c^2 + f^2 ) - ( a^2 + d^2 )\right)^2 \right) \\
K^2 = \frac{1}{16}\left( 4 c^2 f^2 - \left(( a^2 + d^2 ) - ( b^2 + e^2 )\right)^2 \right) 
\end{eqnarray*}$$
where
$$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad d := |BC| \qquad e := |CA| \qquad f := |AB|$$
Then, use $W$, $X$, and $H$ to compute $\angle BC$ via the Law of Cosines, just as you'd use the Law of Cosines for Triangles; likewise, the other dihedral angles.
BTW: I call the above "A Law of Cosines for Tetrahedra", because there's another ...
$$W^2 = X^2 + Y^2 + Z^2 - 2 Y Z \cos \angle OA - 2 Z X \cos \angle OB - 2 X Y \cos \angle OC$$ 
... and in non-Euclidean space, there are even more. For information on this stuff, see my "Hedronometry" page. (I like to think that the level of scholarship improves over time. :)
A: Not a formula but a way to calculate the angle using many in between stages. Maybe later somebody can follow the steps and reduce it into a single formula, I just did not manage that yet.
Assume the vertices of the tetrahedron ( https://en.wikipedia.org/wiki/Tetrahedron ) are named $ A, B, E $ and $F$ where $EF$ is the edge of the dihedral angle and  the angle to calculate is the angle between the faces $\triangle AEF$  and $\triangle BEF$.
The whole method consist of 4 steps:


*

*1a some constructions and calculations on triangle $AEF$ 

*1b some constructions and calculations on triangle $BEF$ 

*2 some calculations on triangle $ABC$

*3 some calculations on triangle $AGC$


1a some constructions and calculations on $\triangle AEF$


*

*let G be the base of the altitude ( https://en.wikipedia.org/wiki/Altitude_%28triangle%29 ) from point of $A$
(so $G$ is on $EF$)

*use the Heron's formula ( https://en.wikipedia.org/wiki/Heron's_formula )  to calculate the  area $O_A$ of  triangle $AEF$ $O_A = \sqrt{s_A(s_A-AE)(s_A-AF)(s_A-EF)}$,
where $s_A = \frac{1}{2}(AE +AF +AE) $


*

*calculate $AG = \frac{O_A}{EF}$

*calculate $EG = \sqrt {AE^2-AG^2}$
1b some constructions and calculations on $\triangle BEF$


*

*let $H$ be the base of the altitude  from point of $B$
(so $H$ is on $EF$)

*let $C$ be the last point to construct rectangle $BHGC$
so $BH$ = $CF$ and $HG$ =$BC$ 


*

*use the Heron's formula  to calculate the  area $O_B$ 


$O_B = \sqrt{s_B(s_B-BE)(s_B-BF)(s_B-EF)}$,
where $s_B = \frac{1}{2}(BE +BF +BE) $


*

*calculate $ CG = BH = \frac{O_B}{EF}$

*calculate $ EH = \sqrt {BE^2-BH^2}$

*calculate $ BC= GH = |EH-EG| $
2 some calculations on $\triangle ABC$
We now can construct the triangle $ABC$
The angle $\angle ACB $ is right so
$ AC = \sqrt {AB^2-AC^2}$
3 finally  $\triangle AGC$
angle $AGC$ is a planar angle we are looking for
using the law of cosine  ( https://en.wikipedia.org/wiki/Law_of_cosines )
we can calculate
$ \angle AGC = \arccos( \frac {AG^2+CG^2-AC^2}{AG \times CG} ) $
And we are done
I was planning to fix it all together in a nice single formula, but was not able to do so yet.
maybe you can try that yourself  (if you manage add it as another answer) 
