Given dihedral angles, find a set of edges In the paper Space Vectors Forming Rational Angles a special set of tetrahedra is mentioned.
"The remaining three are in the R-orbit of the tetrahedron with dihedral angles ($π/7, 3π/7, π/3, π/3, 4π/7, 4π/7$)."
What is a set of edge lengths or vertices for this tetrahedron? I've written a function that converts edges to angles, but I need the reverse.
I solved it in a very messy way: https://community.wolfram.com/groups/-/m/t/2169279. An elegant solution would still be nice.
 A: Let a tetrahedron $OABC$ have edges
$$a := |OA| \qquad b := |OB| \qquad c := |OC| \qquad d := |BC| \qquad e := |CA| \qquad f := |AB|$$
and dihedral angles $A$, $B$, $C$, $D$, $E$, $F$ along respective edges. Let faces $W$, $X$, $Y$, $Z$ be opposite vertices $O$, $A$, $B$, $C$.
The reader is invited to verify these straightforward expressions for edge-lengths in terms of face-areas and volume:

$$(a, b, c, d, e, f) = \frac{2}{3V}\left(
YZ\overline{A}, ZX\overline{B}, XY\overline{C}, WX\overline{D}, WY\overline{E}, WZ\overline{F} \right) \tag{$\star$}$$

where $\overline{\theta} := \sin\theta$. So, "all we have to do" is determine the face-areas and volume. This is not difficult.
Viewing $X$ as the combined shadows of $W$, $Y$, $Z$ (and then viewing $Y$ and $Z$ similarly) gives rise to these analogues of the $a= b\cos C+c\cos B$ relations for a triangle:
$$\begin{align}
X &= W \ddot D + Y \ddot C + Z \ddot B \\
Y &= X \ddot C + W \ddot E + Z \ddot A \\
Z &= X \ddot B + Y \ddot A + W \ddot F
\end{align} \tag1$$
where $\ddot{\theta}:=\cos\theta$. This comprises a simple linear system in $X$, $Y$, $Z$. Solving gives
$$\begin{align}
X \Delta &= W \left(\; \ddot A \left(-\ddot A \ddot D + \ddot B \ddot E + \ddot C \ddot F\right) + \ddot D + \ddot C \ddot E + \ddot B \ddot F \;\right) \\
Y \Delta &= W \left(\; \ddot B \left(\phantom{-}\ddot A \ddot D - \ddot B \ddot E + \ddot C \ddot F\right) + \ddot E + \ddot A \ddot F + \ddot C \ddot D \right) \\
Z \Delta &= W \left(\; \ddot C \left(\phantom{-}\ddot A \ddot D + \ddot B \ddot E - \ddot C \ddot F \right) + \ddot F + \ddot B \ddot D + \ddot A \ddot E \right)\end{align} \tag2$$
where $\Delta := 1 -2\ddot A\ddot B \ddot C - \ddot{A}^2 - \ddot{B}^2 - \ddot{C}^2$. From these, we can calculate volume:
$$V^2 = \frac{2}{9} X Y Z \sqrt{\Delta} \tag3$$
which in turn allows us to find the edge-lengths via $(\star)$. Done! $\square$

For the problem at hand, we take $W=1$ and
$$(A,B,C,D,E,F)=\left(\frac\pi7,\frac\pi3,\frac{4\pi}7,\frac{3\pi}7,\frac\pi3,\frac{4\pi}7\right)$$
then throw everything into Mathematica. The exact trig expressions are a mess, but here are the numerical values:
$$(a,b,c,d,e,f) = (3.1457\ldots, 2.2409\ldots, 2.5227\ldots, 0.9003\ldots, 2.2409\ldots, 2.5227\ldots)$$
Re-scaling so that $a=1$ gives
$$(a,b,c,d,e,f) = (1, 0.7123\ldots, 0.8019\ldots, 0.2862\ldots, 0.7123\ldots, 0.8019\ldots)$$
confirming @Parcly's values.
