[Math] Compute the dihedral angle of a regular pyramid

euclidean-geometrygeometryplatonic-solidssolid-geometry

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.

dihedral angles

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

Best Answer

A vectorial approach would be quite lean and effective.

Dyhedral_angle_1

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

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