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...

Say the four vertices of the shape are $ABCD$, with $\overline{AB} = \overline{CD} = 2$, and the rest of the segments having length $\sqrt{3}$. Let $O$ denote the midpoint of $AB$, then the triangle $AOC$ is a right triangle with hypotenuse $\sqrt{3}$ and one of the legs $1$. We get that $\overline{OC} = \sqrt{2}$. Similarly, $\overline{OD} = \sqrt{2}$. So the triangle $COD$ is an isosceles triangle with two sides of length $\sqrt{2}$ and the third side of length $2$, which implies that it is a right triangle(!).
So one of the dihedral angles is $90^{\circ}$. Now we need to compute the angle between two faces meeting on a side of length $\sqrt{3}$- for example, between  $ABC$ and $DBC$. One way to do this is to assign coordinates to the vertices. From the previous analysis, a particularly nice choice is $A = (-1,0,0)$, $B = (1,0,0)$, $C = (0,\sqrt{2},0)$, $D = (0,0,\sqrt{2})$. 
The plane containing $ABC$ is given by the equation $z = 0$, and the plane containing $DBC$ is given by $x\sqrt{2} + y + z = \sqrt{2}$. Then the angle between the planes is equal to the angle between the normal vectors $\mathbf{n}_1 = (0,0,1)$ and $\mathbf{n}_2 = (\sqrt{2},1,1)$:
$$\cos(\theta) = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{1}{1 \cdot 2} = \frac{1}{2}$$
It follows that the other dihedral angle is $60^{\circ}$. 
