Right n-sided pyramid solid angle Have: 
N-faces right pyramid 
value of apex solid angle
value of N

Need to find:
value of flat angle (C on image)
 A: Hint:
Assuming given apex solid angle is defined in the same way as it is on this Wikipedia entry. Assuming the polygon at the base is regular, so that $\angle AOB = 2 \pi / N$. Assume the radius of the circumcircle of base polygon is 1 (w.l.o.g).
Possible steps to a solution are:


*

*Find $AB$ in the base polygon.

*Find $OF$ in the base polygon.

*Find the height $OS$ of the pyramid using the formula for the solid angle of a right $N$-gonal pyramid with a regular base

*Find $SF$ using $OF$ and $OS$

*Find $c / 2$ using $SF$ and $BF = AB / 2$



The formula referred to on the Wikipedia page for the solid angle of a $n$-gonal right-pyramid with height $h$ and with $r$ as the radius of the circle circumscribing the base is:
$$
\Omega = 2\pi - 2n \arctan\left(\frac {\tan \left({\pi\over n}\right)}{\sqrt{1 + {r^2 \over h^2}}} \right)
$$
Below is an example of the use of this formula for a particular pyramid. Let the base of the pyramid be $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$ and let the apex be $O=(0,0,0)$.
This pyramid occupies one eighth of a full sphere. The solid angle of a full sphere is $4 \pi$ so we expect the solid angle for the pyramid to be $\frac{\pi}{2}$.
The point at the centre of the base is $P=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. The height $h$ of the pyramid is given by $h = |OP| = \frac{1}{\sqrt{3}}$. The radius $r$ of the circumcircle of the base can be obtained from $r = |AP| =\frac{\sqrt{2}}{\sqrt{3}}$. We can then calculate $\sqrt{1 + \frac{r^2}{h^2}} = \sqrt{3}$.
For this pyramid, $n=3$ so $\tan\left(\frac{\pi}{n}\right)= \sqrt{3}$. Putting this together with the above, we can calculate $\Omega$ 
$$
\Omega = 
2\pi - 2n \arctan\left(
\frac{\sqrt{3}}{\sqrt{3}}
\right)
=
2\pi - 2 \times 3 \times \frac{\pi}{4} = \frac{\pi}{2}
$$
confirming the value we expect for the solid angle.
A: Complete formula:
$$t = \frac{tan(\pi/n)^2}{tan(\frac{\Omega-2\pi}{2n})^2}-1$$
$$C = ∠ASB = 2arcsin(\frac{\sqrt t * sin(\pi/n)}{\sqrt {t+1}}) $$ 
