How is it possible to find the dihedral angles of this tetrahedron In this post, the author claims that given a tetrahedron $ABCD$ with known base ABC (i.e. the side lengths $AB$, $BC$, $CA$ are known), and also known angles $\angle ADB, \angle ADC, \angle BDC $, that they can solve for the dihedral angles between the planes of $\triangle ADB$ and $\triangle ADC$, and $\triangle BDC$.

My question is:  How is this possible?  How is it possible to solve for the dihedral angles, without first solving for the side lengths $AD$, $BD$, $CD$?

As suggested in a solution added to the above post, to solve the tetrahedron, I would assign coordinates to the known base triangle $\triangle ABC$.  Then I would define the unknown quantities $x_1, x_2, x_3$ where $x_1 = DA , x_2 = DB, x_3 = DC $ and apply the law of cosines to the various lateral triangles, which would result in three quadratic equations in $x_1, x_2, x_3$.  These equations take the form
$ \dfrac{x_1^2 + x_2^2 - AB^2 }{2 x_1 x_2} = \cos \angle ADB $
$ \dfrac{x_1^2 + x_3^2 - AC^2 }{2 x_1 x_3} = \cos \angle ADC $
$ \dfrac{x_2^2 + x_3^2 - BC^2 }{2 x_2 x_3} = \cos \angle BDC $
These three equations can be solved numerically for $x_1, x_2, x_3$.
The next step is to find the coordinates of vertex $D$.  So let $D = (d_1, d_2, d_3) $, then it follows that
$ (a_1 - d_1)^2 + (a_2 - d_2)^2 + (a_3 - d_3)^2 = x_1^2$
$ (b_1 - d_1)^2 + (b_2 - d_2)^2 + (b_3 - d_3)^2 = x_2^2$
$ (c_1 - d_1)^2 + (c_2 - d_2)^2 + (c_3 - d_3)^2 = x_3^2$
These equations are easy to solve due to their special structure, where taking the differences between the first and second equations and the first and the third equations, results in two planar equations whose intersection gives a line along which the point $(d_1, d_2, d_3)$ lies.  Substituting in one of the original three equations, gives two possible coordinates for the vertex $D$.
Once the coordinates of $D$ are known, then we have solved the tetrahedron completely and all the sides and angles and dihedral angles can be computed in a straight forward manner.
 A: As it happens, the "face-angles" surrounding a tetrahedral vertex are all you need to know to determine the dihedral angles at that vertex (and vice versa). What happens at the vertex doesn't care about the opposite face.
The connection is made through what's called the "law of cosines for spherical geometry", although that terminology makes the results seem more exotic and scary than they really are. They follow from straightforward (if slightly tedious) Euclidean trig or vector methods.
Given face-angles $\alpha := \angle BDC$, $\beta := \angle CDA$, $\gamma := \angle ADB$, and writing simply $A$, $B$, $C$ for dihedral angles along edges $DA$, $DB$, $DC$, the relations are
$$\begin{align}
\cos\alpha &= \cos\beta\cos\gamma+\sin\beta\sin\gamma\cos A \\
\cos\beta &= \cos\gamma\cos\alpha+\sin\gamma\sin\alpha\cos B \\
\cos\gamma &= \cos\alpha\cos\beta +\sin\alpha\sin\beta\cos C 
\end{align}$$
As a sanity check: Consider a corner of a room where two walls and meet the floor. The floor need not be square; let's say there's a face-angle $\alpha$ between the two wall-floor edges. We'll assume the walls themselves are upright in the sense that they meet along an edge that makes right face-angles with the wall-floor edges (so, $\beta=\gamma=90^\circ$). Substituting this information into the above relations gives
$$\begin{align}
\cos\alpha &= \cos90^\circ\cos90^\circ+\sin90^\circ\sin90^\circ\cos A \;\quad\to\quad &\cos\alpha &= \cos A\\
\cos90^\circ&= \cos90^\circ\cos\alpha\;\;+\sin90^\circ\sin\alpha\;\;\,\cos B 
\;\quad\to\quad &0 &= \cos B\\
\cos90^\circ&= \;\;\cos\alpha\cos90^\circ+\;\sin\alpha\;\sin90^\circ\;\cos C 
\,\quad\to\quad &0&=\cos C
\end{align}$$
These confirm that the walls are upright in another sense, making right dihedral angles with the floor: $B=C=90^\circ$. Moreover, the dihedral angle between the walls exactly matches the face-angle on the floor: $A=\alpha$.
These relations have counterparts that switch the roles of face- and dihedral angles (and change one sign, so be careful!):
$$\begin{align}
\cos A &= -\cos B\cos C+\sin B\sin C\cos \alpha \\
\cos B &= -\cos C\cos A+\sin C\sin A\cos \beta \\
\cos C &= -\cos A\cos B+\sin A\sin B\cos \gamma
\end{align}$$
Note that substituting dihedral angle information from the corner of our room ($B=C=90^\circ$) would give us the face-angle information ($\beta=\gamma=90^\circ$, $\alpha=A$).
Importantly, these calculations require no knowledge of the shape of the rest of the room. In the tetrahedral context: edge-lengths are unnecessary; face-angles alone determine dihedral angles, and vice-versa.
(Of course, edge-lengths can help you find any missing face-angles you might need, but this isn't the scenario described in the question.)
