I have an equation to calculate $A$ from $r$ and $h$: $$ A= \arccos\left(\frac{r-h}{r}\right)r^2 - (r-h)\sqrt{r^2-(r-h)^2} $$
But now $A$ and $r$ are known, and I want to solve for $h$ instead. So far, I solve this equation numerically, but I was wondering whether it can be solved analytically. All three $A$, $r$ and $h$ are positive. I have tried to solve it using Wolfram Alpha, but I only get a timeout. Any better ideas?
Assume $r>0$ and let $h=r(1-x)$. This gives $$\cos ^{-1}(x)-x \sqrt{1-x^2}=\frac A {r^2}=\color{red}{k}$$ Use a series expansion around $x=0$ $$k=\frac \pi 2-2x+\sum_{n=1}^\infty \frac {|a_n|}{b_{n+1}}\,x^{2n+1}\tag 1$$ where the $a_n$ form sequence $A091154$ in $OEIS$ and the $b_n$ form sequence $A143582$.
Truncate to some order and use power series reversion
$$x=-\frac 18 \sum_{n=1}^\infty \frac {c_n}{4^{n} (2 n-1)!}\,\left(k-\frac{\pi }{2}\right)^{2n-1}$$ where the $c_n$ form sequence $A281181$ in $OEIS$.
To give an idea of the accuracy of $(1)$, consider the truncated series $$S_p=\frac \pi 2-2x+\sum_{n=1}^p \frac {|a_n|}{b_{n+1}}\,x^{2n+1}$$ and the infinite norm $$\Phi_p=\int_{-1}^{+1} \Big(\cos ^{-1}(x)-x \sqrt{1-x^2} -S_p\Big)^2\,dx$$
$$\left( \begin{array}{cc} p & \Phi_p \\ 1 & 1.19707\times 10^{-3} \\ 2 & 1.85427\times 10^{-4} \\ 3 & 5.21327\times 10^{-5} \\ 4 & 1.99086\times 10^{-5} \\ 5 & 9.16556\times 10^{-6} \\ 6 & 4.78667\times 10^{-6} \\ 7 & 2.73743\times 10^{-6} \\ 8 & 1.67653\times 10^{-6} \\ 9 & 1.08327\times 10^{-6} \\ 10 & 7.30691\times 10^{-7} \\ \end{array} \right)$$
