An irregular tetrahedron

Let $OABC$ be the tetrahedron, $ABC$ its base (the equilateral triangle), $h=AD$ the required height, $E$ the midpoint of $AB$ and $DF\bot BC$ .
The plane $OEC$ is a plane of symmetry of the tetrahedron, hence $D$ lies on $CE$ . Moreover, by the Theorem of three perpendiculars, $OF\bot BC$ and $OE\bot AB$ .

Let $OAB$ be the lateral face that makes the angle of $80{}^\circ$ with the base. Then, $\angle OED=80{}^\circ$ and $\angle OFD=60{}^\circ$ .
In $\triangle OED$ , ${{a}_{1}}=ED=h\cdot \cot 80{}^\circ \ \ \ \ \ (1)$

In the equilateral $\triangle ABC$ , the median $CE$ is also angle bisector of $\angle ACB$ , hence, $\angle DCF=30{}^\circ$ , so the right triangles $\triangle ADF$ and $\triangle CDF$ are congruent.
Thus, ${{a}_{2}} \coloneqq DC=OF\coloneqq {{l}_{1}}$ .

In $\triangle ODF$ , $DF={{l}_{1}}\cdot \cos 60{}^\circ \Rightarrow DF=\dfrac{{{l}_{1}}}{2}\Rightarrow DF=\dfrac{{{a}_{2}}}{2}$ .

By Pythagoras’ theorem on $\triangle ODF$ , $O{{D}^{2}}=O{{F}^{2}}-D{{F}^{2}}\Rightarrow {{h}^{2}}={{a}_{2}}^{2}-{{\left( \frac{{{a}_{2}}}{2} \right)}^{2}}\Rightarrow {{a}_{2}}=\dfrac{2}{\sqrt{3}}h \ \ \ \ \ (2)$ .

In the equilateral $\triangle ABC$ we have $AE=BC\cdot \frac{\sqrt{3}}{2}\Rightarrow AE=5\sqrt{3} \ \ \ \ \ (3)$ .

Combining $(1)$ , $(2)$ and $(3)$ , we get $\begin{aligned} CE={{a}_{1}}+{{a}_{2}} & \Rightarrow 5\sqrt{3}=h\cdot \cot 80{}^\circ +\dfrac{2}{\sqrt{3}}h \\ & \Rightarrow h=\dfrac{15}{2+\sqrt{3}\cdot \cot 80{}^\circ } \\ & \Rightarrow h\approx 6.50644251 \\ \end{aligned}$

For the answer, $\left\lfloor 1000h \right\rfloor =\boxed{6506}$ .