Maximum number of circles tangent to two concentric circles

Depending on what is meant by non-overlaping circles, there is another possible way to get an answer of $3$.

You don't need to compute $\arcsin \frac45$ exactly, because you're going to be taking the floor anyway.

You just need to know that the angle opposite the side of length $4$ in a $(3,4,5)$ right triangle is somewhere between $45^\circ$ (the threshold between $3$ and $4$ circles) and $60^\circ$ (the threshold between $2$ and $3$ circles).

You can figure this out by comparing the sides $(3,4,5)$ to the sides $(s,s,\sqrt2 s)$ and $(s, \frac{\sqrt3}{2}s, 2s)$ that we'd see in triangles with a $45^\circ$ and $60^\circ$ degree angle, respectively.

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 $\displaystyle 4>\frac{360^\circ}{\theta}>3$.