Find the edge angle of a dodecahedron using spherical trigonometry?

Question

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!

Answer 1

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?

Answer 2

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.

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

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 $\color{red}{\sin(a)=\frac{2}{3}}$.