How to express this expression in terms of N? I am trying to express this formula in terms of N:
$$
A=\frac{a^2N} {a\cdot \tan{\frac{180}{N}}}
$$
I really don't know how to do this. I tried and got this:
$$
A \cdot a \cdot \tan{\frac{180}{N}} = a^2N
$$
$$
\tan{\frac{180}{N}}=\frac{a^2N}{Aa}
$$
$$
\frac{180}{N}=\tan^{-1}{\frac{a^2N}{Aa}}
$$
$$
N=\frac{180}{\tan^{-1}{\frac{a^2N}{Aa}}}
$$
But I just can't get the N on the right side. The $\tan$ is so annoying! How do I do this?

Equations which mix polynomial and trigonometric terms do not show analytical solutions and numerical methods should be involved.
Let us consider the equation $$A=\frac{a^2N} {a\cdot \tan({\frac{\pi}{N})}}=\frac{aN} {\tan({\frac{\pi}{N})}}$$ So, we can consider that we look for the zero of the function $$f(N)=\frac{aN} {\tan({\frac{\pi}{N})}}-A$$ $$g(N)={aN} -A\,{\tan({\frac{\pi}{N}})}$$ $$h(N)=a N \cos(\frac{\pi}{N})-A \sin(\frac{\pi}{N})$$ Probably, the last equation will be better to consider since discontinuties have been removed.
For illustration purposes, let us consider the case $a=2$, $A=1$. A plot of the function $h(N)$ shows a solution somewhere between $2$ and $3$. So, let us start Newton method using $N_0=2$; this will generates the following iterates $$N_1=2.31830988618379$$ $$N_2=2.31348787275960$$ $$N_3=2.31348827662319$$ which is the solution for fifteen significant figures.
If we consider the case where $N$ could be large (this would correspond to large values of ratio $\frac A a$, Taylor expansion would give $$h(N)=a N-\frac{\frac{\pi ^2 a}{2}+\pi  A}{N}+O\left(\frac{1}{N^2}\right)$$ that is to say $$N\approx\sqrt{\frac{\pi }{2}} \sqrt{\frac{\pi  a+2 A}{a}}$$ In the case considered above, this would lead to $N\approx \frac{1}{2} \sqrt{\pi  (2+2 \pi )}\approx 2.55061$ which is not too bad.
Using $a=2$, $A=35$, the estimate would be $7.74033$ while the solution would be $7.63580$.
Edit
You could be amazed by the fact that, using a $[2,2]$ Pade approximant (do not worry : you will learn about them sooner or later), the solution can be approximated by $$N=\frac 12 \sqrt{\frac \pi {3a}}\sqrt{2 A \left(\frac{2 A}{\pi  a+2 A}+5\right)+5 \pi  a}$$ which, for the two worked cases gives $\approx 2.34191$ and $\approx 7.63777$.
Still more funny, using a $[2,4]$ Pade approximant, the solution can be approximated by $$N= \frac 15 \sqrt{\frac \pi {6a}}\sqrt{2 A \left(\frac{4 A (37 \pi  a+84 A)}{5 \pi ^2 a^2+20 \pi  a A+24
   A^2}+61\right)+61 \pi  a}$$  which, for the two worked cases gives $\approx 2.31681$ and $\approx 7.63584$.
