# 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? 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 $$\sin\delta=\cos(\beta/2)/\cos(\alpha/2)$$?

• I’m not able to access the page that you’ve cited just now, but I’d guess that $F$ and $G$ are defined to be the feet of the altitudes to $D$, creating right angles by definition.
– amd
Feb 13, 2018 at 22:28
• @amd the plane GFD is defined to be perpendicular to the line AC of intersection of two planes CAD and CBD. I'm sure you're right that implies DG is an altitude of BDC, but for some reason I'm blanking. Actually I guess what I want to see is that DG is perpendicular to plane BAC. So DG is perpendicular to both BC and FG. Feb 14, 2018 at 0:12
• I don't think you have formulated the problem with enough information. For example, consider a square pyramid, and let $s = 1.$ Now you have the number of sides and $s,$ but if the square is a square of side $1/4$ you get very different dihedral angles than if the square has side $1.$ The formula you cite at the end is derived for regular polyhedra, where $\alpha$ is determined by the type of polyhedron and therefore the edge of the regular polygon, $2s \sin\frac\alpha2,$ is also determined. If you meant the polygon to have side $1,$ that needs to be part of the problem statement. Feb 14, 2018 at 2:30
• @DavidK I think the formula $2s\sin(\alpha/2)$ for the side opposite $\alpha$ should apply to any isosceles triangle. I don't think I have assumed equilateral triangular faces or regular polyhedra anywhere in the equations quoted. If the dihedral angle depends on the ratio of the slant edge to the base edge, I will be happy to accept that. Feb 14, 2018 at 2:58
• The issue is not whether the side has length $2s \sin\frac\alpha2$ -- of course it does! The issue is that in your formulation, we cannot find the value of of $2s \sin\frac\alpha2$, whereas in the page you linked to, that value can easily be found, because that page most definitely _does assume it is dealing with a regular polyhedron. On the other hand, if the polygon's side is $1,$ then $s$ is the ratio of the slant edge to the base edge, and we're both happy. Feb 14, 2018 at 4:19

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 $$\cos \alpha = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}}$$

• One thing that's awkward about my solution is that I have to talk about perpendicularity of lines that do not actually intersect. I'm sure there's a right way to talk about those. Vectors solve this problem. Feb 14, 2018 at 21:35
• @ziggurism: to my knowledge, vectorial approach is the most effective: You can easily calculate all angles, lengths and surfaces of interest. Feb 14, 2018 at 21:39
• Yes of course that's true, but if a geometry problem is presented in a synthetic way (Euclidean geometry style), I like to solve it that way. Feb 14, 2018 at 21:42
• @ziggurism: ok, I only wish you do not want to use also ..the greek numeral system. Feb 14, 2018 at 23:20
• Let's compromise: roman numerals? Feb 15, 2018 at 2:43

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:

1. 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 $$\cos\gamma = \frac{\sin\left(\frac\alpha2\right)}{\sin\left(\frac\beta2\right)}\tag{1}\label{1}.$$

2. 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 $$\sin\delta=\frac{\cos\gamma\sin\beta}{\sin\alpha} \stackrel{\eqref{1}}= \frac{\sin\left(\frac\alpha2\right)\cdot2\sin\left(\frac\beta2\right)\cos\left(\frac\beta2\right)}{\sin\left(\frac\beta2\right)\cdot2\sin\left(\frac\alpha2\right)\cos\left(\frac\alpha2\right)}=\frac{\cos\left(\frac\beta2\right)}{\cos\left(\frac\alpha2\right)}\tag{2}\label{2}.$$

3. 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.

• Very helpful answer, but I would just say there is a complication where $\beta > 90^\circ$ for then $G$ no longer lies on line segment $BC$ but is located out to the left of $B$, eg in the case of the tetrahedron where $\beta = 120^\circ$. Jan 4 at 17:17

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 : $$\begin{eqnarray*} 2 \sin^2 \frac{1}{2} \delta -1 & = & \frac{\sin^2 \gamma}{\cos^2 \frac{1}{2} \alpha} \cdot \cos \beta + \cos^2 \gamma \cdot \frac{\cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\ & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \left( 1 - \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot (\cos^2 \frac{1}{2}\beta - \sin^2 \frac{1}{2}\beta) + \left( \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot \cos^2 \frac{1}{2} \beta \right] \\ & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + \sin^2 \frac{1}{2} \alpha \right] \\ \Rightarrow 2 \sin^2 \delta & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + 1 \right] \\ & = & \frac{2 \cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\ \Rightarrow \mbox{since } \alpha, \beta, \gamma \in (0, 180^\circ), \hspace{1.5em} \sin \delta & = & \frac{\cos \frac{1}{2} \beta}{\cos \frac{1}{2} \alpha}. \end{eqnarray*}$$