<s>[INST] trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles? What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...
 [/INST] Here's a hedronometric approach ...
Let $a$ and $d$ be the (necessarily opposite) edges of length $2$. These edges are orthogonal to each other in space; moreover, their projections into a mutually-parallel plane form the diagonals of a square of side-length $\sqrt{2}$. This square is a "pseudo-face" of the tetrahedron, and has area $H = 2$.
The other pairs of opposite edges ---say, $b$ & $e$ and $c$ & $f$--- determine the remaining two pseudo-face areas $J$ and $K$. Although the tetrahedron's symmetry dictates that these areas must be equal, the values are a little harder to intuit than $H$. However, the Sum of Squares formula tells us that 
$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$
where $W$, $X$, $Y$, $Z$ are the areas of the tetrahedron's (ordinary) faces. In this case, by Heron's Formula, $W = X = Y = Z = \sqrt{2}$. We deduce that $J = K = \sqrt{2}$.
Finally, if we assume that faces $Y$ and $Z$ meet along edge $a$, bounding dihedral angle $A$, and that $Z$ and $X$ meet along edge $b$, bounding dihedral angle $B$, then the tetrahedral Law of Cosines says 
$$\begin{align}
H^2 &= Y^2 + Z^2 - 2 Y Z \cos A \qquad\to\qquad \cos A = \,\,0 \qquad\to\qquad A = \frac{\pi}{2} \\[4pt]
J^2 &= Z^2 + X^2 - 2 Z X \cos B \qquad\to\qquad \cos B = \frac{1}{2} \qquad\to\qquad B = \frac{\pi}{3}
\end{align}$$
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