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

Dodecahedron

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

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

$\frac{\mathrm{15}+7\sqrt{5}}{4}\bullet {a}^{3}$

$\mathrm{sin}\left(\pi /5\right)=$ $\sqrt{\frac{5-\sqrt{5}}{8}}$

Divide the circle into 120 divisions wthout transendental formulas $\mathrm{16}\bullet \mathrm{sin}\left(\pi /\mathrm{60}\right)=$ $2\bullet \left(1-\sqrt{3}\right)\bullet \sqrt{5+\sqrt{5}}+\sqrt{2}\bullet \left(\sqrt{5}-1\right)\bullet \left(\sqrt{3}+1\right)$

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