<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] Let $\alpha$ be the vertex angle of a face, and $\beta$ a base angle. Then, by the Law of Cosines: 
$$\cos\alpha = \frac{-2^2 + \sqrt{3}^2 + \sqrt{3}^2}{2\cdot\sqrt{3}\sqrt{3}} = \frac{1}{3} \qquad \cos\beta = \frac{2^2 - \sqrt{3}^2 + \sqrt{3}^2}{2\cdot 2\sqrt{3}} = \frac{\sqrt{3}}{3}$$
Also,
$$\sin\alpha = \frac{2\sqrt{2}}{3} \qquad \sin\beta = \frac{\sqrt{6}}{3}$$
Each vertex of your tetrahedron is surrounded by one $\alpha$ angle and two $\beta$ angles. Let $A$ be the dihedral angle opposite face angle $\alpha$ at a vertex; and let $B$ be opposite a $\beta$. By the Spherical Law of Cosines, we can calculate
$$\begin{align}
\cos A &= \frac{\cos\alpha-\cos^2\beta}{\sin^2\beta} = \frac{1/3-1/3}{2/3} = 0 \\[4pt]
\cos B &= \frac{\cos\beta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta} = \frac{\sqrt{3}/3\,(1-1/3)}{2\sqrt{12}/9} = \frac{1}{2}
\end{align}$$
Therefore, 

$$A = \frac{\pi}{2} \qquad\qquad B = \frac{\pi}{3}$$

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