Find the altitude of a tetrahedron whose faces are congruent triangles Any set of four identical triangles can be arranged to form a four-surface solid. If these four triangles are not equilateral, e.g. 3,4,5, how do I find the altitude of from the one used for the bottom to the apex formed by the other three.

Elaborating on my comment ...
Let the triangle have side-lengths $a$, $b$, $c$, and area $T$. The Cayley-Menger determinant for area ---equivalently Heron's formula--- tells us
$$16 T^2 = (-a+b+c)(a-b+c)(a+b-c)(a+b+c) \tag{1}$$
The CM determinant for volume gives
$$288V^2 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & a^2 & b^2 & c^2 \\
1 & a^2 & 0 & c^2 & b^2 \\
1 & b^2 & c^2 & 0 & a^2 \\
1 & c^2 & b^2 & a^2 & 0
\end{matrix}\right| = 4 (-a^2 + b^2 + c^2) (a^2 - b^2 + c^2) (a^2 + b^2 - c^2) \tag{2}$$
Observe that, if the triangle has a right angle, then $(2)$ vanishes; the tetrahedron has zero volume. Moreover, if the triangle has an obtuse angle, then $(2)$ is negative, making $V$ imaginary: in this case, there must be no tetrahedron at all.
Since altitude $h$ satisfies $3 V = h T$,

$$h^2 = \frac{9 V^2}{T^2} = \frac{2(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)(a+b+c)} \tag{$\star$}$$

For something more angle-related ... By the Law of Cosines, we have
$$a^2 = b^2 + c^2 - 2 b c \cos A \qquad\text{etc}$$
so that
$$V^2 = \frac{1}{72}\cdot 2 b c \cos A \cdot 2c a \cos B \cdot 2 a b \cos C = \frac{1}{9} a^2 b^2 c^2 \cos A \cos B \cos C$$
By the Law of Sines,
$$a = d \sin A \qquad\text{etc}$$
where $d$ is the circumdiameter of the triangle, so that
$$T = \frac12 a b \sin C = \frac12 d^2 \sin A \sin B \sin C \qquad
V = \frac13 d^3\sin A \sin B \sin C\sqrt{\cos A \cos B \cos C} $$
Hence,

$$h = \frac{3V}{T} = 2d\sqrt{\cos A \cos B \cos C} \tag{$\star\star$}$$

