<s>[INST] Maximum number of circles tangent to two concentric circles In a recent math contest, the following question arose:

Two concentric circles of radii 1 and 9 make a ring. In the interior of this ring $n$ circles are drawn without overlapping, each being tangent to both of the circles of the ring. What is the largest possible value for $n$?

I solved it like this:
The radius of each of the small circles must be $(9-1)/2=4$. I connected the centres of two of the small circles together, and also to the centre of the large circle. Let the central angle be $\theta$. The triangle formed has side lengths $4+1=5$, $4+1=5$, and $4+4=8$. This can be split in two to get two 3-4-5 triangles. Since these triangles are right, we can solve for $\theta$:
$$
\begin{align}
\sin\frac{\theta}{2}&=\frac{4}{5}\\
\frac{\theta}{2}&=\arcsin{\frac{4}{5}}\\
\theta&=2\arcsin\frac{4}{5}.
\end{align}
$$
Now the answer to the question is simply $\left\lfloor\frac{2\pi}{\theta}\right\rfloor=3$.
However, this contest was a no calculator contest, and thus I was not able to compute $\arcsin(4/5)$ for the answer. How do you solve this question without a calculator?
 [/INST] Note that $\displaystyle \frac{\sqrt{2}}{2}<\frac{4}{5}<\frac{\sqrt{3}}{2}$. We have $\displaystyle \sin45^\circ<\sin\frac{\theta}{2}<\sin60^\circ$.
$90^\circ<\theta<120^\circ$.
Hence $\displaystyle 4>\frac{360^\circ}{\theta}>3$.
</s>