Height of an irregular tetrahedron with an equilateral base and lateral faces making angles $60^\circ$, $60^\circ$, $80^\circ$ with that base 
An irregular tetrahedron has a base that is an equilateral triangle of side length $10$.  The lateral faces make angles of $60^\circ, 60^\circ$ and $80^\circ$ with the base.  Find the height of the tetrahedron.

So, one way I thought I could solve this problem is using coordinate geometry, specifically attaching a reference frame to the base, and writing the equations of the three planes that represent the three lateral faces, and then solving the linear system for the apex coordinates.


Let $ABC$ denote the equilateral triangle where the two dihedral angles with values 60° taken along $AB$ and $AC$. Let $H$ be the midpoint of $BC$
and $D$ be the apex of the tetrahedron. Vertical plane $HAD$ is clearly a plane of symmetry of this tetrahedron.
Let $F$ be the foot of the altitude dropped from $D$ onto the horizontal plane; $F$ belongs clearly to line $AH$. We have to find $f:=\text{length}(DF)$.
Let $d:=\text{length}(FH)$.
First, we have:
$$\frac{d}{h}=c \ \ \text{where} \ \ c:=\cot(80°)\approx 0.1763269 \tag{1}$$
It is known (see for example page 3 of this document) that the barycentric coordinates of $F$ with respect to base triangle $ABC$ are proportional to the products of the different sidelengths of the base triangle by the cotangent of their associated dihedral angles:
$$(\text{length(AB)} \cot(60°), \ \ \ \text{length(AC)} \cot(60°),  \ \ \  \text{length(BC)} \cot(80°)),$$
themselves proportional to:
$$(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},c) \tag{2}$$
The "true" barycentrical coordinates being obtained by dividing expressions (2) by their sum: $c+\frac{2}{\sqrt{3}}$.
Using the classical areal interpretation of these "true" barycentric coordinates
and taking into account the fact that the area of triangle $ABC$ is $25 \sqrt{3}$, one can identify 2 different ways to express the ratio $\frac{area(HBC)}{area(ABC)}$, i.e.,
$$\frac{\tfrac12 d \times 10}{25 \sqrt{3}}=\frac{c}{c+\frac{2}{\sqrt{3}}}\tag{3}$$
From (3), one deduces
$$d=\dfrac{5 \sqrt{3}c}{c+\frac{2}{\sqrt{3}}}$$
Then, from (1), we get the value of
$$h=\frac{d}{c} \approx 6.5064425...$$
