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.

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. :)