Dihedral Angle of an Icosidodecahedron

If you extend the faces of all the pentagons of the icosidodecahedron, it will form a dodecahedron where each vertex is centered above one of the equilateral triangles.

A pentagon in the icosidodecahedron can be formed by joining the midpoints of a pentagon in the dodecahedron. Since the icosidodecahedron pentagon has sides of $1$ , the dodecahedron pentagon has sides of $s = 2 \cdot \frac{1}{2} \sec 36° = \frac{1}{2}(\sqrt{5} - 1)$ .

Each of the "extra" $20$ pyramids formed by the difference of the dodecahedron and icosidodecahedron have equilateral triangle bases with sides of $1$ and lateral edges of $l = \frac{1}{2}s = \frac{1}{4}(\sqrt{5} - 1)$ . The apothem of the equilateral triangle base is $a = \frac{1}{2\sqrt{3}}$ , and the radius is $r = \frac{1}{\sqrt{3}}$ , so by the Pythagorean Theorem the height of the pyramid is $h = \sqrt{l^2 - r^2} = \sqrt{(\frac{1}{4}(\sqrt{5} - 1))^2 - (\frac{1}{\sqrt{3}})^2} = \frac{1}{6}(3\sqrt{3} - \sqrt{15})$ .

The dihedral angle between the base and lateral face of the pyramid is then $\theta_1 = \tan^{-1} \frac{h}{a} = \tan^{-1} \frac{\frac{1}{6}(3\sqrt{3} - \sqrt{15})}{\frac{1}{2\sqrt{3}}} = \tan^{-1} (3 - \sqrt{5})$ . That means the dihedral angle between a pentagonal face and an equilateral triangular face of the icosidodecahedron is $\theta_2 = 180° - \theta_1 = 180° - \tan^{-1} (3 - \sqrt{5}) \approx \boxed{142.62°}$ .