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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.

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This is not an answer, just something that might help: You can calculate the vertex angles from the edge lengths and the solid angles from the vertex angles, and then you have four linear equations relating the four solid angles to the six vertex angles, so only two degrees of freedom left to be determined -- but perhaps you already knew all that... – joriki Jul 4 '11 at 4:35

2 Answers 2

up vote 4 down vote accepted

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. :)

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Thanks for that, the solution is different (in the sense that it is original) to other approaches I have seen online. Have you extended this technique to higher dimensions? – Kyle Jul 5 '11 at 0:33
@Kyle: I developed this Hedronometry stuff independently, though I'm sure at least some of it exists elsewhere. It's well-known, for instance, that the cell-based Pythagorean Theorem generalizes in a straightforward way to simplices in any-dimensional space ("square of content of hypotenuse-cell = sum of squares of contents of leg-cells"), as does the Law of Cosines in the end of my answer. There don't seem to be such clean counterparts of the Law of Cosines from the beginning of my answer, though, which hinders investigation (or even definition) of appropriate pseudo-elements. – Blue Jul 5 '11 at 3:20

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 ( ) 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 $\triangle AEF$

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 ( )

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)

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