Alternative methods for deriving the internal angle in a tetrahedron.

Question

The internal angle $\theta$ in a regular tetrahedron is $109.47^\circ$.

I can derive this using the following method and am now looking for alternative methods.

Take a regular tetrahedron with vertices $ABCD$ and centre $E$. The internal angle that I am referring to is of course any of the $\angle$ vertex-centre-vertex permutations such as $\angle AEB$.

The tetrahedron is composed of $4$ congruent triangular based pyramids, $ABCE$, $ABDE$, $ACDE$, and $BCDE$.

The height of each pyramid with $E$ as the apex is $h$ and the area of each base face is $A$.

The height of the regular tetrahedron is $(h+l)$ and the area of it's base face is also $A$.

The volume of the total tetrahedron is four times the volume of each pyramid.

Therefore: $$\frac{1}{3}A(h+l)=4 \frac{1}{3}Ah$$

This reduces to: $$l=3h$$

$$\cos \theta = -\frac{1}{3}$$

$$\theta = 109.47^\circ$$

I'm now looking for alternative methods for finding this internal angle.

Answer 1

Blue has already provided a convenient coordinate representation of a regular tetrahedron in a comment: $x,y,z=\pm1$ with $xyz=1$. All we have to do is choose two vertices, e.g. $\pmatrix{1\\1\\1}$ and $\pmatrix{-1\\-1\\1}$, and calculate the cosine using dot products:

$$ \cos\theta=\frac{\pmatrix{1\\1\\1}\cdot\pmatrix{-1\\-1\\1}}{\sqrt{\pmatrix{1\\1\\1}\cdot\pmatrix{1\\1\\1}}\sqrt{\pmatrix{-1\\-1\\1}\cdot\pmatrix{-1\\-1\\1}}}=-\frac{1}3\;. $$

Answer 2

Let $BH$ be an altitude of our tetrahedron $ABCD$ and $AB=1$.

Thus, $$BH=\sqrt{AB^2-AH^2}=\sqrt{1-\left(\frac{1}{\sqrt3}\right)^2}=\sqrt{\frac{2}{3}},$$ which gives $$AE=BE=\frac{3}{4}BH=\frac{\sqrt3}{2\sqrt2}$$ and we obtain: $$\measuredangle AEB=2\arcsin\frac{\frac{AB}{2}}{BE}=2\arcsin\sqrt{\frac{2}{3}}\approx109.47^{\circ}.$$