How to find the dihedral angle of shapes in higher dimensions? The dihedral angle of a tetrahedron (e.g side length of 1) is quite straightforward, splitting up into smaller triangles and eventually using simple trigonometry, we get 70.5 degrees. However, whilst searching for a similar method for the 5-cell (4 dimensional analogue of a tetrahedron), the calculations relied on:
$arccos$ ($d^{-1}$), where d is the dimension of the shape.
For example for the tetrahedron, we get $arccos$ ($3^{-1}$) which of course is 70.5 degrees. This only works with the simplex series. Where does this formula come from ?
How do we go about calculating the dihedralangles of 4d shapes (let's focus on just regular polytopes, and preferably avoid using vectors)
 A: For a regular simplex in $d-$space, look at the centroid $G$ of an arbitrary $d-2$ face $s_{d-2}$. Furthermore, there is a pair of vertices, call them $V_i$ and $V_j$, which are from the total $d$ simplex, but not from the $d-2$ face $s_{d-2}$
(they are what's left of the vertices of the $d$ simplex when you remove all the vertices of the $d-2$ face $s_{n-2}$). This centroid is the center of the $d-2$ face $s_{d-2}$, which itself is a $d-2$ regular simplex obtained as the intersection of
the pair of adjacent $d-1$ regular simplicial faces $V_i\,s_{d-2}$ and $V_j\,s_{d-2}$ of the $d$ simplex. Because
the $d$ simplex is regular, all of its faces are also regular simplices of various dimensions.
Hence, each segment $A_iG$ and $A_jG$ is perpendicular to the
$d-2$ affine subspace spanned by the $d-2$ simplex $s_{d-2}$. Moreover, these two segments are equal in length, i.e.
$V_iG = V_jG = h_{d-1}$ because they are the heights of two congruent regular simplices of dimension $d-1$, namely $V_is_{d-2}$ and $V_js_{d-2}$.
Therefore, the triangle $V_iV_jG$ is isosceles, where $V_iG = V_jG = h_{d-1}$ and $V_iV_j=1$.
The dihedral angle you are interested in is then $$\angle \, V_iGV_j \, = \, \theta$$

If you denote by $G_i$ and $G_j$ the centroids (centers) of the $d-1$ simplices
$V_i  \, s_{d-2}$ and $V_j\,s_{d-2}$ respectively, then $G_i \in V_iG$ and $G_j \in V_jG$, as well as
$V_iG_j \, \perp \, V_jG$ and $V_jG_i \, \perp \, V_iG$.
The segments $V_iG_j$ and $V_jG_i$ are congruent heights in the
isosceles triangle $V_iV_jG$ and the triangle $V_iG_jG$ is right-angled, so
$$\cos(\theta) \, =\, \cos(\angle \, V_iGV_j) \, =\, \cos(\angle \, V_iGG_j) \, =\, \frac{G_jG}{V_iG}$$
and by the reflection symmetry of the isosceles triangle $V_iV_jG$, the two segments $G_iG = G_jG$ so
$$\cos(\theta) \,  =\, \frac{G_iG}{V_iG}$$
Observe that in the latter equality, $V_iG$ is the height of the regular $d-1$ simplex $V_i\,s_{d-2}$ that connects the vertex $V_i$ with the center $G$ of the
opposing regular $d-2$ simplex $s_{d-2}$. Moreover, the point $G_i$ is the center of the $d-1$ simplex $V_i \, s_{d-2}$.
Now everything boils down to calculating the ratio in which the center $G_i$ of a regular $d-1$ simplex $V_i\,s_{d-2}$ splits the height $V_iG$ connecting $V_i$
with the center of the regular $d-2$ simplex $s_{d-2}$:
We will do this by showing how it works in one dimension higher ($d-1$ regular face of a $d$ regular simplex, instead of the current $d-2$ regular face of a $d-1$ regular simplex)
and use it as an induction step
in $d$, calculating the ratio in which the center $G_d$ of the total regular simplex $s_{d}$ splits the height $V_iG_j$. In the current notations,
$G_d = V_iG_j \, \cap \, V_jG_i$. The triangles $V_iG_iG_d$ and $V_iG_jG$ are right-angled and similar, so
$$\frac{G_jG}{V_iG} = \frac{G_iG_d}{V_iG_d} = \cos(\theta)$$
But by the fact that the traigle $V_iV_jG$ is isosceles, $$G_jG = G_iG \,\,\,\, \text{ and } \,\,\,\, G_iG_d = G_jG_d$$
which yields the ratio
$$\frac{G_iG}{V_iG} = \frac{G_jG_d}{V_iG_d} = \cos(\theta)$$
Let me flip it
$$\frac{V_iG_d}{G_dG_j} = \frac{V_iG}{GG_i} = \frac{1}{\cos(\theta)}$$
I claim that $$\frac{V_iG_d}{G_dG_j} = d$$
The ratio $\frac{V_iG}{GG_i}$ is a ratio that comes from the analogous situation one dimension lower, i.e.
the position of the canter $G_i$ of the $d-1$ regular simplex
$V_i \, s_{d-2}$ on the height $V_iG$ of this same simplex. If, by induction, we assume that it has been shown that
$$\frac{V_iG_i}{G_iG} = d-1$$ then $$\frac{V_iG}{GG_i} = \frac{V_iG_i + G_iG}{G_iG} = \frac{V_iG_i}{G_iG} + 1 = d - 1 + 1 = d$$
Hence
$$\frac{V_iG_d}{G_dG_j} = \frac{V_iG}{GG_i} = d = \frac{1}{\cos(\theta)}$$ and there you have that
$$\cos(\theta) = \frac{1}{d}$$ and consequently the dihedral angle you are after is
$$\theta = \arccos\left(\frac{1}{d}\right)$$
The beginning of the inductive step can start with $d=2$, in which case we have that we have an equilateral triangle, where the ratio is known to be $\frac{1}{2}$.
