Height of a tetrahedron How do I calculate the height of a regular tetrahedron having side length $1$ ?
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?

You can also use trig based on the dihedral angle between two faces of the tetrahedron.
Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$
$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.
As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:
$$\begin{eqnarray}
|OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\
s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\
s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\
1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\
\frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\
\frac{1}{3} &=& \cos{M} \;\;\; (**)\\
\Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M}
\end{eqnarray}$$
Therefore,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$
(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):
$$\begin{eqnarray}
\cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\
\cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\
\cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n}
\end{eqnarray}$$
(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)
