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"?
 A: 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
$$2\pi -10\arctan\bigg(\frac{\tan(\frac{\pi}{5})}{\sqrt{1+r^2/h^2}}\bigg)\approx 0.29507$$
A: Based on the website provided by N74, I am posting the calculations that can obtain the answer to the question in my post. I would appreciate the efforts of someone knowledgeable in solid geometry to peruse my calculations and tell me whether it is right.
I got an answer of about 4.7866. If the solid angle of the whole sphere is $4\pi$, this answer is too big, considering that there are 12 black regular pentagons on a soccer ball.
My calculations
I am looking to calculate the measure of the solid angle subtended by each pentagon of a soccer ball.
$a$ denotes the edge length of each regular pentagon. According to
math.wikia.org/wiki/Pentagonal_pyramid,

the altitude of each pyramid with a regular pentagonal base is
$$h = a\sqrt{\frac{1}{2} - \frac{1}{2}\sqrt{\frac{1}{5}}} \approx 0.52573 a.$$
According to
wikipedia.org/wiki/Truncated_icosahedron,

the radius of the sphere circumscribing the truncated icosahedron - the soccer ball - is
$$r = \frac{a}{4}\sqrt{58 + 18\sqrt{5}} \approx 2.478 a.$$
According to
https://en.m.wikipedia.org/wiki/Solid_angle,

the measure of the solid angle subtended by each pentagon is
$$2\pi - 2(5)\arctan\sqrt{\frac{\tan(\pi/5)}{\sqrt{1 + r^2/h^2}}}.$$
As
$$\sqrt{1 + r^2/h^2} \approx 4.818358,$$
the measure of the solid angle subtended by each pentagon is approximately 4.7866.
A: In the following required quantities are defined and arrived at by direct calculation using only the Sine/Cosine Laws in spherical trigonometry.
Areas of a spherical triangle $ = ( A+B+C-\pi)\; R^2 $ is dimensioned spherical excess. The spherical excess in turn is nothing but the solid angle$ =  A+B+C-\pi. $  The three concepts of area, spherical excess and solid angle are bound together. The solid angle has also been called Integral Curvature by Gauss.
Examples: For a full sphere of area  $4 \pi R^2$  a solid angle $4 \pi$. A spherical triangle with three right angles has solid angle $\pi/2$ like an octant of a sphere. A hemisphere can be considered to be enclosed by a spherical triangle with 3 vertices $120^{\circ}$ apart on equator, so spherical excess is $ 3 \pi -\pi= 2 \pi. $
In what follows it is assumed that all pentagon/ hexagon boundary lines of the soccer ball (formed by spherical icosahedron truncation) are part of great circles..i.e., the center of the soccer ball, P and Q lie in an equatorial plane. Euclidean length of the icosahedron edge is 1. A correction is applied for circular arc. We consider the corner spherical triangle $OPQ$ of the icosahedron shown magnified in the hand sketch below to base calculations.

The side of central curvilinear hexagon is $h$. It is required to calculate $h, \beta$ before Area of pentagon and Solid angle are calculated.
Arc length $s$ of the curved icosahedral side is slightly larger than edge length $1$;
$$ s= 2 R_{sph}\sin^{-1}({1}/{2 R_{sph}})\tag1 $$
Law of Cosines supplies an implicit relation to for numerical evaluation:
$$ \cos h = \cos ^2 [(s-h)/2] + \sin ^2 [(s-h)/2] \cos 2 \pi/5 \tag2$$
Law of Sines to find $ \beta $
$$ \dfrac{\sin [(s-h)/2]}{\sin \beta}= \dfrac{\sin h}{\sin 2\pi/5} \tag3 $$
Area of a spherical triangle (non-dimensionalized) by definition/formula is its spherical excess.. $ 5 \times \Delta OPQ :$
$$ Area= 5( \;2 \pi/5 + 2 \beta-\pi)= 10 \beta - 3\pi \tag4$$
Definition of solid angle is the non-dimensionalized Area.
$$ Solid Angle=\dfrac{Area\; on \;Sphere}{R_{sph}^2} \tag5 $$
Analytical expressions with surds etc. avoided in numerical work, can be inserted if it is felt that it would simplify.
Numerical results respectively approx.
$$( R_{sph},s, h,\beta, Surface Area, Solid \;Angle )$$
$$=(0.951057,1.05296,0.386778,0.969114,0.266365,0.294486) \tag6 $$
Solid Angle $= 0.294486 $ steradians subtended at sphere center by the convex pentagon.
A: 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
$$\boxed{\color{red}{\omega_p}=\color{blue}{2\pi-10\sin^{-1}\left(\frac{9-\sqrt5}{12}\right)\approx 0.295072263}\ sr}$$
