Area between arc and line I have been working on a problem for a few days now. This was a challenge problem on a lecture for Trigonometry. I managed to find an equation for the radius, but wasn't able to solve it.
Problem: Line $AB$ is drawn such that $\overline{AB} = 20$. Minor arc $AB$ is drawn with endpoints $AB$ such that the length of arc $AB$ is $21$. Find the area of the region bounded by the arc and the line.
Progress: First, draw the full diagram as shown below.

We can compute the area of $\angle{ABC}$. Using the Law of Cosines, we have that
\begin{align*}
\cos C &= \frac{a^2+b^2-c^2}{2ab} \\
&= \frac{2r^2-400}{2r^2}\\
&= 1-\frac{200}{r^2}
\end{align*}Therefore, $\angle ACB = \cos^{-1}\left(1-\frac{200}{r^2}\right)$. The length of the arc in terms of $\angle ACB$ is
\begin{align*}
&\frac{\angle ACB}{360}2\pi r \\
&= \frac{\angle ACB}{180}\pi r \\
&= \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r
\end{align*}We know that
\begin{align*}
&21 = \frac{\cos^{-1}\left(1-\frac{200}{r^2}\right)}{180}\pi r 
\implies \\&\frac{3780}{\pi r} = {\cos^{-1}\left(1-\frac{200}{r^2}\right)} 
\implies\\ &\cos\left(\frac{3780}{\pi r}\right) = 1-\frac{200}{r^2}
\end{align*}
However, I couldn't solve this equation. Any help would be appreciated. I can also visualize a calculus approach involving finding the area between two curves, but I want to solve it in an elementary way if possible.
 A: hint
Let $t $ be the angle $\angle ACB .$
Then
$$21=rt $$
and
$$r\sin(t/2)=10 .$$
the area we want is the difference
$$S=\frac {r^2}{2}t-\frac{20.h}{2} $$
with
$$h=r\cos(t/2) $$
hence

$$S=\frac {21}{2}r-10\sqrt {r^2-100} $$

with $r $ satisfying
$$r\sin (\frac {21}{2}r)=10$$
A: As angryavian commented, the problem reduces to : find the zero of $$f(\theta)=\frac \theta {\sin\left(\frac\theta 2\right)}- 2.1\qquad \text{or}\qquad g(\theta)=\theta-2.1{\sin\left(\frac\theta 2\right)}$$ which can only be solved using numerical methods.
Let $x=\frac\theta 2$ to make the equation $$x=1.05 \sin(x)$$ and use, for an approximation, the magnificent $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ was proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).
Skipping the trivial $x=0$, this leads to the quadratic 
$$4 x^2+\left(\frac{84}{5}-4 \pi \right) x+\pi\left(5 \pi -\frac{84  }{5}\right)=0$$ the positive solution of which being $$x=-\frac{21}{10}+\frac{\pi }{2}+\frac{1}{40} \sqrt{7056+3360 \pi -1600 \pi ^2}\approx 0.537445$$ making $$\theta \approx 1.07489$$ while the exact solution would be $\approx 1.07682$.
With this first result, we could use a Taylor expansion around $\theta=\frac \pi 3$ and get $$g(\theta)=\left(\frac{\pi }{3}-\frac{21}{20}\right)+\left(1-\frac{21 \sqrt{3}}{40}\right)
   \left(\theta-\frac{\pi }{3}\right)+\frac{21}{160} \left(\theta-\frac{\pi
   }{3}\right)^2+O\left(\left(\theta-\frac{\pi }{3}\right)^3\right)$$ Ignoring the higher order terms, another quadratic leading to $$\theta \approx 1.07683$$
