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.

Given a tetrahedron with base $ABC$ and apex $D$, let

*

*$E$ be the orthogonal projection of $D$ onto the plane holding $ABC$.

*$h = |DE|$ will be the height of tetrahedron.

*$\theta_A / \theta_B / \theta_C$ be the angle between faces $DBC$ / $DCA$ / $DAB$ and base $ABC$.

*$\ell_A / \ell_B / \ell_C$ be the distance of $E$ to edges $BC$ / $CA$ / $AB$.

As long as all $\theta_A, \theta_B, \theta_C < 90^\circ$, $E$ lies inside $ABC$.
Furthermore, we have

*

*$\ell_A = h \cot\theta_A$, $\ell_B = h \cot \theta_B$ and $\ell_C = h\cot\theta_C$

*$|BC|\ell_A + |CA|\ell_B + |BC|\ell_C = 2\verb/Area/(ABC)$
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,
$$\begin{align}
h = \frac{\ell_A + \ell_B + \ell_C}{\cot\theta_A + \cot\theta_B + \cot\theta_C}
&= \frac{5\sqrt{3}}{2\cot(60^\circ) + \cot(80^\circ)} = \frac{15}{2 + \sqrt{3}\cot(80^\circ)}\\
&\sim 6.506442514261543\end{align}$$
