Dodecahedronal, Icosahedronal Symmetry: The cosines of the angle between adjacent face normals are $cos(∙) = \frac{1}{\sqrt{5}}$ and $cos(∙) = \frac{\sqrt{5}}{3}$ respectively

abstract: At exact_exp.xml the pertenant dodecahedron and icosahedron lenghts, which are derived at proof_new.txt, are exhibited. Further derivation is at plato.txt, and the ratios for the dodecahedron inscribed sphere radius equal to one are exhibited on this page. For deriving the volume of the dodecahedron, it is easier and more direct to have these values first. Further, it is shown how to know if a form of [a+b[5]]/[c] can be reduced to a form (d[5]+e)/f.

Pentagon

Dodecahedron

edge length	$\sqrt{50 - 22 \sqrt{5}}$	1	$\frac{\sqrt{12 + 4 \sqrt{5}}}{\sqrt{6}}$
inscribed	1	$\frac{\sqrt{25 + 11 \sqrt{5}}}{\sqrt{40}}$	$\frac{\sqrt{5 + 2 \sqrt{5}}}{\sqrt{15}}$
center of edge radius	$\frac{\sqrt{5 - \sqrt{5}}}{\sqrt{2}}$	$\frac{3 + \sqrt{5}}{4}$	$\frac{\sqrt{3 + \sqrt{5}}}{\sqrt{6}}$
superscribed	$\sqrt{15 - 6 \sqrt{5}}$	$\frac{\sqrt{9 + 3 \sqrt{5}}}{\sqrt{8}}$	1
pentagon height	$\frac{\sqrt{15 - 5 \sqrt{5}}}{\sqrt{2}}$	$\frac{\sqrt{5 + 2 \sqrt{5}}}{2}$	$\frac{\sqrt{5 + \sqrt{5}}}{\sqrt{6}}$
pentagon area	∙	$\frac{\sqrt{25 + 10 \sqrt{5}}}{4}$	∙
total surface area	$30 ∙ \sqrt{2} ∙ \sqrt{65 - 29 \sqrt{5}}$	$3 ∙ \sqrt{5} ∙ \sqrt{5 + 2 \sqrt{5}}$	∙
volume	$10 ∙ \sqrt{2} ∙ \sqrt{65 - 29 \sqrt{5}}$	$\frac{\sqrt{5} \sqrt{47 + 21 \sqrt{5}}}{2 \sqrt{2}}$	∙

wiki/Exact_trigonometric_constants#Uses_for_cnstants Volume, where a is the length of an edge

\frac{15 + 7 \sqrt{5}}{4} ∙ a^{3}

\sin (π / 5) =

\sqrt{\frac{5 - \sqrt{5}}{8}}

Divide the circle into 120 divisions wthout transendental formulas $16 ∙ \sin (π / 60) =$ $2 ∙ (1 - \sqrt{3}) ∙ \sqrt{5 + \sqrt{5}} + \sqrt{2} ∙ (\sqrt{5} - 1) ∙ (\sqrt{3} + 1)$

For pi/7 angles the solution is transcendental but for pi/60 the solution is irrational