Another question asks for the dihedral angle of a general tetrahedron in terms of its edge lengths. I chose to answer by giving a formulas in terms of face areas (because those relations deserve to be better-known). Most-relevant here is this version, for a tetrahedron $OABC$, ...
$$W^2=X^2+Y^2+Z^2-2YZ\cos\angle OA - 2ZX\cos\angle OB-2XY\cos\angle OC \tag{$\star$}$$
where $W := |\triangle ABC|$, $X := |\triangle OBC|$, $Y := |\triangle AOC|$, $Z := |\triangle ABO|$ are face areas, and $\angle OA$, $\angle OB$, $\angle OC$ are dihedral angles along respective edges $\overline{OA}$, $\overline{OB}$, $\overline{OC}$.
In a regular tetrahedron, all face areas are equal ($W=X=Y=Z$), as are all dihedral angle measures ($\angle OA = \angle OB = \angle OC$), so $$W^2 = 3 W^2 - 6W^2\cos\angle OA \qquad\to\qquad \cos\angle OA = \frac{1}{3} \tag{$\star\star$}$$
I'll leave calculating the angle measure itself to the reader. $\square$
FYI ... The $0$-dimensional point, $1$-dimensional segment, $2$-dimensional triangle and $3$-dimensional tetrahedron have higher-dimensional counterparts, each known simply by the umbrella term "$n$-dimensional simplex" (or "$n$-simplex"). The bounding elements (counterparts of "vertices", "edges", and "faces") are called "facets", and we can write, for $n\geq 2$ ...
Fun Fact. The angle between two neighboring facets of a regular $n$-simplex is $\operatorname{arccos}\frac{1}{n}$ .
So, always remember and never forget: $\cos 60^\circ$ isn't merely "half of one"; it's the reciprocal of the equilateral triangle's dimension!
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.)
Best Answer
Given a tetrahedron with base $ABC$ and apex $D$, let
As long as all $\theta_A, \theta_B, \theta_C < 90^\circ$, $E$ lies inside $ABC$. Furthermore, we have
For the tetrahedron at hand, we have $|AB| = |BC| = |CA| = 10$ and $\verb/Area/(ABC ) = \frac{\sqrt{3}}{4}(10)^2$.
This leads to
$$\ell_A + \ell_B + \ell_C = 5\sqrt{3}$$
and as a result, $$\begin{align} h = \frac{\ell_A + \ell_B + \ell_C}{\cot\theta_A + \cot\theta_B + \cot\theta_C} &= \frac{5\sqrt{3}}{2\cot(60^\circ) + \cot(80^\circ)} = \frac{15}{2 + \sqrt{3}\cot(80^\circ)}\\ &\sim 6.506442514261543\end{align}$$