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

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