<s>[INST] Measure of an angle "subtended by each pentagon" in a truncated icosahedron A soccer ball is a truncated icosahedron; it consists of 12 black regular pentagons and 20 white regular hexagons; the edge lengths of the pentagons and hexagons are congruent. What is the measure of the solid angle (in ste-radian) subtended by each pentagon of the soccer ball?
My Explanation:
If the edge length of the regular pentagon is $a$, according to mathworld.wolfram.com, the radius of the circle circumscribing the pentagon is
\begin{equation*}
R = \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a \approx 0.8507 a ,
\end{equation*}
and the radius of the circle inscribed in the pentagon is
\begin{equation*}
r = \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a \approx 0.6882 a.
\end{equation*}
So, the length of the line segment between a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is
\begin{equation*}
R + r
= \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a + \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a
\approx 1.5388 a.
\end{equation*}
According to wikipedia.org/wiki/Truncated_icosahedron, the radius of the sphere circumscribing the truncated icosahedron is 
\begin{equation*}
\frac{a}{4} \sqrt{58 + 18\sqrt{5}} \approx 2.478 a.
\end{equation*}
An implementation of Heron's Formula yields the distance between the center of the sphere and the midpoint of a side of the pentagon of about $2.427a$. An implementation of the Law of Cosines shows that the angle with its vertex at the center of the icosahedron and its endpoints at a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is approximately $36.5^\circ$.
Is that correct? Is that what is meant by "the angle subtended by each pentagon"?
 [/INST] We know that the normal distance $H_p$ of each of 12 regular pentagonal faces from the center of a truncated icosahedron with edge length $a$ is given by following formula
$$H_p=\frac{a(2\sqrt5+1)}{\sqrt{10-2\sqrt5}}=\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}$$
The solid angle subtended by any regular polygonal plane with each side $a$ at any point lying on the perpendicular at a distance $h$ from the center is given by generalized formula from HCR's Theory of Polygon as follows
$$\omega=2\pi-2n\sin^{-1}\left(\frac{2h\sin\frac{\pi}{n}}{4h^2+a^2\cot^2\frac{\pi}{n}}\right)$$
Now, substituting the corresponding values $h=H_p=\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}$ and $n=5$ in above formula, the solid angle subtended by each regular pentagon $\omega_p$ at the center of truncated icosahedron (soccer ball) is
$$\omega_p=2\pi-2\cdot 5\sin^{-1}\left(\frac{2\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}\sin\frac{\pi}{5}}{4\left(\frac{a}{4}\sqrt{\frac{250+82\sqrt5}{5}}\right)^2+a^2\cot^2\frac{\pi}{n}}\right)$$
After simplifying, one will get
$$\boxed{\color{red}{\omega_p}=\color{blue}{2\pi-10\sin^{-1}\left(\frac{9-\sqrt5}{12}\right)\approx 0.295072263}\ sr}$$
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