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How can I find the edge angle (the angle at the center of a polyhedron subtended by an edge of the polyhedron) of a dodecahedron (a polyhedron with 3 pentagonal faces meeting at each vertex)?

I know how to find the edge angle of a regular polyhedron when its faces are triangles - I envision one of the faces on the surface of a sphere and since the triangles are all equal, I can easily use the Law of Cosines.

However, I am very confused about what to do if the face is a pentagon - what would I do with a pentagon on the surface of a sphere?

Thank you!

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Take the center of the pentagon and connect it to each of its vertices. You get five congruent triangles. Can you determine their internal angles? Can you go on from there?

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    $\begingroup$ Right. I think you can find all three angles of one of these, and use the all-angles Law of Cosines. $\endgroup$ – Lubin Apr 4 '16 at 22:39
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Can you obtain 3D Cartesian coordinates for the vertices of the unit dodecahedron (distance from center to all vertices == 1, with the origin [ 0 0 0 ] at the center of the enclosing sphere)?

Can you select two vertices that share a common edge? (call them v1 and v2)

If so, then you can use this theorem from direction vector geometry:

d1d2 = cos (theta)

The first step is to recognize that, if the origin is at the center of the solid, then the vertices v1, v2... are equivalent to direction vectors d1, d2... because they all have unit lengths.

Do you know how to calculate the vector dot product in 3D?

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Firstly we need the relationship between the vertex and edge angle for a regular spherical N-gon. Lets call the edge angle $a$ and the vertex angle $2 \theta$. Now draw geodesics from the center of the N-gon to two adjacent verticies ... this is illustrated in the brilliantly drawn digram below.

enter image description here Now the cosine & sine rules yield \begin{eqnarray*} \cos(a)=\cos^2(r)+\sin^2(r) \cos( \phi)= 1 - 2 \sin^2(r) \sin^2 \left( \frac{\phi}{2} \right) \\ \frac{\sin(a)}{\sin{\phi}} = \frac{\sin(r)}{\sin{\theta}} \Rightarrow \sin^2(a) \sin^2(\theta) = 4 \sin^2(r) \sin^2 \left( \frac{\phi}{2} \right) \cos^2 \left( \frac{\phi}{2} \right) \end{eqnarray*} Now rearrange the cos rule, multiply by $2\cos^2 \left( \frac{\phi}{2} \right)$, we have \begin{eqnarray*} 2(1-\cos(a))\cos^2 \left( \frac{\phi}{2} \right) &=& 4 \sin^2(r) \sin^2 \left( \frac{\phi}{2} \right) \cos^2 \left( \frac{\phi}{2} \right) = \sin^2(a) \sin^2(\theta)\\ 1+\cos(a) &=& \frac{2 \cos^2 \left( \frac{\phi}{2} \right)}{\sin^2(\theta)} \end{eqnarray*} Remembering that $\phi=\frac{2 \pi}{N}$, the above formula gives the relation between the edge angle $a$ and vertex angle $2 \theta$. The special case $N=4$ can be found here ... Trigonometric rule on a spherical square

enter image description here

REF WIKI: https://en.wikipedia.org/wiki/Spherical_polyhedron#Examples For this specific question we have $N=5$ and we require $3$ pentagons to meet at a vertex, recall the special values \begin{eqnarray*} \sin(\theta)= \sin \left( \frac{ \pi}{3} \right) = \frac{ \sqrt{3}}{2} \\ \cos\left( \frac{ \phi}{2} \right) = \cos\left( \frac{ \pi}{5} \right) = \sqrt{ \frac{3+\sqrt{5}}{8}}. \end{eqnarray*} After a little numerical algebra we have the side length of spherical dodecahedron is $\color{red}{\cos(a)= \frac{\sqrt{5}}{3}}$ or $\color{red}{\sin(a)=\frac{2}{3}}$.

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