<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] The wikipedia about solid angle says that 

The solid angle of a right $n$-gonal pyramid, where the pyramid base is a regular $n$-sided polygon of circumradius $r$, with a pyramid height $h$ is
  $$2\pi -2n\arctan\bigg(\frac{\tan(\frac{\pi}{n})}{\sqrt{1+r^2/h^2}}\bigg)$$

Note here that our $r$ is the circumradius of the pentagon (not the radius of the sphere circumscribing the soccer ball). So, according to the wikipedia about regular pentagons, we get
$$r=a\sqrt{\frac{2}{5-\sqrt 5}}=a\sqrt{\frac{5+\sqrt 5}{10}}$$
Now, considering a right triangle $OAB$ where $O$ is the center of the soccer ball, $A$ is the center of the pentagon and $B$ is a vertex of the pentagon, we get
$$\small h=OA=\sqrt{OB^2-AB^2}=\sqrt{\bigg(\frac{a}{4}\sqrt{58+18\sqrt 5}\bigg)^2-\bigg(a\sqrt{\frac{5+\sqrt 5}{10}}\bigg)^2}=a\sqrt{\frac{41+25\sqrt 5}{8\sqrt 5}}$$
So, the solid angle we seek is
$$2\pi -10\arctan\bigg(\frac{\tan(\frac{\pi}{5})}{\sqrt{1+r^2/h^2}}\bigg)\approx 0.29507$$
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