Compute the dihedral angle of a regular pyramid Given a regular pyramid, defined as a right pyramid with a base which is a regular polygon, with the vertex above the centroid of the base, I would like to compute the dihedral angle between adjacent faces, as well as the angle between a lateral face and the base, in terms of the number of sides and the slant edge length. Can someone walk through the calculation?
Using Google I found this helpful article about computing dihedral angles by Greg Egan. 

We have isosceles triangle $CAD$ above the plane of triangle $CBD$, making a dihedral angle of $\epsilon$. Triangle $CBD$ is the orthogonal projection of $CAD$, so also isosceles. The apex angle between adjacent edges in triangle $CAD$ is $\alpha$. The angle of triangle $CBD$ is $\beta.$ Edge $AC$ makes a slant angle of $\gamma$ with the projected edge $BC$.
There is a plane perpendicular to slant edge $AC$ through the opposite vertex $D$, which intersects edge $AC$ at point $F$, and $BC$ at point $G$. Angle $GFD$ is $\delta.$
If triangle $CAD$ is the side of a regular pyramid, and $CBD$ its base, then $2\delta$ will be the dihedral angle between adjacent faces, and $\epsilon$ will be the dihedral angle between slant face and base.
I can see that plane $DFG$ is perpendicular to slant edge $AC$, hence why triangles $AFD$ and $AFG$ are right triangles. But why are $DGC$ and $FGD$ right angles?
How do I arrive at $$\sin\delta=\cos(\beta/2)/\cos(\alpha/2)$$?
 A: The short answer is, $DG$ lies in the plane $BCD$ perpendicular to $AB$, and is therefore (parallel to a line) perpendicular to $AB$. $DG$ also lies in the plane $DFG$ perpendicular to $AC$, and therefore is (parallel to a line) perpendicular to $AC$. Being perpendicular to both $AB$ and $AC$, $DG$ is perpendicular to every line in the plane $ABC$ that it intersects, including both $FG$ and $BC$.

Let's walk through the derivation of all three formulas in detail. Following the linked article, we see:


*

*First, from the slant face isosceles triangle $\triangle CAD$, using the right triangle with altitude $AE,$ and $AC=s,$ we have $EC=s\sin(\alpha/2).$
Altitude $AB$ is perpendicular to plane $CAD,$ so triangle $ABC$ is a right triangle, and we have $BC=AC\cos\gamma=s\cos\gamma.$ From the base isosceles triangle $\triangle CBD$ we have $EC=s\cos\gamma\sin(\beta/2).$ 
Equating the two expressions gives $$\cos\gamma = \frac{\sin\left(\frac\alpha2\right)}{\sin\left(\frac\beta2\right)}\tag{1}\label{1}.$$

*Next, why are $FG$ and $BC$ (that is, plane $ABC$) perpendicular to $DG$? 
The trick is to realize that the perpendicular to a plane through the vertex of a triangle is orthogonal to all lines in that plane, including the two edges that meet it, as well as (a parallel line to) the opposite edge.
By construction, plane $DFG$ is perpendicular to $AC$. Therefore $AC$ is perpendicular to any line in $DFG$, ergo $AC$ is perpendicular to $FG$ and $FD$. 
Additionally, line $DG$ lies in plane $BCD,$ so it (or a line parallel to it) is perpendicular to $AB$, as $AB$ is perpendicular to plane $BCD.$  Line $DG$ is also in plane $DFG,$ so perpendicular to $AC$. Therefore it is perpendicular to plane $ABC.$ And therefore also to $FG.$ Angles $\angle DGF$ and $\angle DGC$ are right angles (line $BC$ is also in plane $ABC$ and so perpendicular to $DG$).
Therefore looking at right triangle $\triangle FGD$ we have $GD=s\sin\alpha\sin\delta.$
And looking at right triangle $\triangle BGD$ we see $GD=s\cos\gamma\sin\beta.$
Equating gives $$\sin\delta=\frac{\cos\gamma\sin\beta}{\sin\alpha} \stackrel{\eqref{1}}= \frac{\sin\left(\frac\alpha2\right)\cdot2\sin\left(\frac\beta2\right)\cos\left(\frac\beta2\right)}{\sin\left(\frac\beta2\right)\cdot2\sin\left(\frac\alpha2\right)\cos\left(\frac\alpha2\right)}=\frac{\cos\left(\frac\beta2\right)}{\cos\left(\frac\alpha2\right)}\tag{2}\label{2}.$$

*Looking at right triangle $\triangle ABE$ we have $BE=AE\cos\epsilon.$ 
From triangle $\triangle AEC$ we have $AE=s\cos\left(\frac\alpha2\right).$
And from triangle $\triangle BEC$ we have $BE=s\cos\left(\gamma\right)\cos\left(\frac\beta2\right).$
Thus $$\cos\epsilon=\frac{\sin\left(\frac\alpha2\right)\cos\left(\frac\beta2\right)}{\cos\left(\frac\alpha2\right)\sin\left(\frac\beta2\right)}=\frac{\tan\left(\frac\alpha2\right)}{\tan\left(\frac\beta2\right)}\tag{3}\label{3}.$$
So given the vertex angle $\alpha$ and the projection of that angle into the base $\beta$, we have the edge angle $\gamma$, the dihedral angle $\delta$ between $ABC$ and $ACD$, and the dihedral angle $\epsilon$ between $ACD$ and $BCD$.
Applying this to the regular pyramid, where $CAD$ is a lateral face and $\triangle ABC$ is a vertical (perpendicular to the base) cross section through an edge between adjacent faces, then the dihedral angle between adjacent side faces is $2\delta.$
In terms of the number of sides of the pyramid $n$, since there are $n$ angles of measure $\beta$ around a vertex in the plane, we have $$\beta=\frac{2\pi}{n}.$$ And in terms of the slant edge length $AC=s$, and base side length $CD=\ell$, we have $$\ell=2s\sin\left(\frac\alpha2\right).$$ So in terms of $n,\ell,$ and $s,$ we write $$\begin{align}\cos\gamma&=\frac{\ell}{2s\sin\left(\frac{\pi}{n}\right)},\\\sin\delta&=\frac{2s\cos\left(\frac{\pi}{n}\right)}{\sqrt{4s^2-\ell^2}},\\\cos\epsilon&=\frac{\ell}{\tan\left(\frac{\pi}{n}\right)\sqrt{4s^2-\ell^2}}.\end{align}$$
For example, putting $n=4$, $s=\ell$ gives us the regular square pyramid with $\gamma=\frac{\pi}{4},$ $\sin\delta=\sqrt{\frac{2}{3}},$ and $\cos\epsilon=\frac{1}{\sqrt{3}}.$ Which matches the angles of the regular octahedron in the table of Platonic solid dihedral angles. 
Or with $n=3$, $s=\ell$ we get $\cos\gamma=1/\sqrt{3}$ and $\cos(2\delta)=\cos\epsilon=1/3,$ again agreeing with known geometry of the regular tetrahedron.
For a non-Platonic example, a regular hexagonal pyramid with height $h=\sqrt{3}\ell$ and slant edge length $s=2\ell$ will have $\cos(2\delta)=-3/5$, the angle of a $3$-$4$-$5$ triangle.
A: A vectorial approach would be quite lean and effective.

Given two faces of the pyramid, sharing the common edge  $V P_n$,
and containing the contiguous base points $P_{n-1}$ and $P_{n+1}$,
the dihedral angle between these two faces would be the angle
made by the two vectors ($t_m, t_p$), normal to the common edge and
lying on the respective face.
Clearly, that will be also the angle made by the normal vectors to the faces,
provided that one is taken in the  inward, and the other in the outward
direction.
That is, by the right-hand rule,
$$
{\bf n}_{\,m}  = \mathop {P_{\,n} P_{\,n - 1} }\limits^ \to  \; \times \;\mathop {P_{\,n} V}\limits^ \to  \quad \quad {\bf n}_{\,p}  = \mathop {P_{\,n} P_{\,n + 1} }\limits^ \to  \; \times \;\mathop {P_{\,n} V}\limits^ \to  
$$
Then the dihedral angle $\alpha$ will be simply computed from the dot product
$$
\cos \alpha  = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}}
$$
A: An alternate approach to obtaining the formula for $\delta$ is to take a unit normal $\underline{\mathbf{u}}$ for a given slope and then rotate this by the angle $\beta$ about the $z$-axis, ie about the vector $\underline{\mathbf{k}}$, to produce a unit normal $\underline{\mathbf{v}}$ for the adjacent slope, as shown in Fig. 1. This approach avoids the complication which arises when $\beta > 90^\circ$, for then $G$ is no longer on the line segment $BC$, but extends out to the left beyond $B$ (eg. in the case of the tetrahedron we have $\alpha = 60^\circ$ and $\beta = 120^\circ$, giving dihedral angle $2\delta = \arccos (1/3)$).

Since $\underline{\mathbf{u}}$ and $\underline{\mathbf{v}}$ are both outward pointing, we have :
$$
\underline{\mathbf{u}} \cdot \underline{\mathbf{v}} = \cos(180 - 2\delta) = -\cos 2\delta.
$$
From Fig. 1 we have :
$$
\underline{\mathbf{u}} = \sin \epsilon \; \underline{\mathbf{i}} + \cos \epsilon \; \underline{\mathbf{k}}.
$$
Rotating $\underline{\mathbf{u}}$ about the $z$-axis by angle $\beta$ leaves the $\underline{\mathbf{k}}$ component unchanged and changes the component $\sin \epsilon \; \underline{\mathbf{i}}$ to $\sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \; \underline{\mathbf{j}}$ so that :
$$
\underline{\mathbf{v}} = \sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \;\underline{\mathbf{j}} + \cos \epsilon \; \underline{\mathbf{k}}
$$
Thus :
\begin{eqnarray}
2 \sin^2 \frac{1}{2} \delta -1 = -\cos 2\delta & = & \underline{\mathbf{u}} \cdot \underline{\mathbf{v}} \nonumber \\
& = & \sin^2 \epsilon \cos \beta + \cos^2 \epsilon \label{eq:Rodrigues} \tag{1}
\end{eqnarray}
We require an expression that involves $\alpha$ and $\beta$ only. Denote the altitude of the regular pyramid by $h$, and the radius of its regular polygonal base by $r$. Then :
\begin{equation}
\sin \epsilon = \frac{h}{s \cos \frac{1}{2} \alpha}, \hspace{1em} \mbox{and} \hspace{1em} \cos \epsilon = \frac{r \cos \frac{1}{2} \beta}{s \cos \frac{1}{2} \alpha} \label{eq:angle-epsilon} \tag{2}
\end{equation}
with :
\begin{equation}
\frac{h}{s} = \sin \gamma \hspace{1em} \mbox{and} \hspace{1em} \frac{r}{s} = \cos \gamma. \label{eq:angle-gamma} \tag{3}
\end{equation}
As in the article we can readily establish by computing the length of EC in two ways that :
\begin{equation}
\cos \gamma = \frac{\sin \frac{1}{2} \alpha}{\sin \frac{1}{2} \beta}. \label{eq:angle-gamma-formula} \tag{4}
\end{equation}
Thus putting equations (\ref{eq:Rodrigues}) - (\ref{eq:angle-gamma-formula}) together we now have :
\begin{eqnarray*}
2 \sin^2 \frac{1}{2} \delta -1 & = & \frac{\sin^2 \gamma}{\cos^2 \frac{1}{2} \alpha} \cdot \cos \beta + \cos^2 \gamma \cdot \frac{\cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\
& = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \left( 1 - \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot (\cos^2 \frac{1}{2}\beta - \sin^2 \frac{1}{2}\beta) + \left( \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot \cos^2 \frac{1}{2} \beta \right] \\
& = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + \sin^2 \frac{1}{2} \alpha \right] \\
\Rightarrow 2 \sin^2 \delta & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + 1 \right] \\
& = & \frac{2 \cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\
\Rightarrow \mbox{since } \alpha, \beta, \gamma \in (0, 180^\circ), \hspace{1.5em} \sin \delta & = & \frac{\cos \frac{1}{2} \beta}{\cos \frac{1}{2} \alpha}.
\end{eqnarray*}
