I have the question "What angle is subtended at the center of a circle of radius $2$ km by an arc of length $9$ m?"
I am not sure which formula to use to find the subtended angle.
[Math] Finding subtended angle at the centre of a circle from known arc length
anglearc lengthcircleseuclidean-geometrygeometry
Related Solutions
The length of the arc is proportional to the angle by proposition 6.33 in Euclid's Elements. We can choose units to make the constant of proportionality equal to $1$ for a circle of radius $1$. Those units are called radians.
[I give the Euclid reference since it uses the notion of angle that might be familiar to you. Nowadays we find it more economical to define the trignometric functions without any appeal to geometry (for eg. via the complex exponential), then define angles by inverting one of these functions, and then derive your result as an exercise in calculus.
Here is a simple hand-waving argument along those lines. Let OA and OB be two radii of a circle of radius $r$. Then by simple trignometry the length of the line segment AB is $2r\sin(\theta/2)$. For small enough $\theta$, $\sin(\theta/2) \approx \theta/2$ so the length of AB is approximately $r\theta$. In this case the length of AB is also close to the length of the arc AB hence that is also approximately $r\theta$. We can make the approximation as good as we like by making $\theta$ small enough. We can handle large arcs by approximating it by a large number of small segments of this sort.]
Here a way to solve your problem: in the attached figure you can calculate the theta angle in the right triangle $\triangle{XX'O}$ which is the half of the isosceles triangle $\triangle{XOY}$ with angle $\angle{XOY}=2\alpha\lt 60^{\circ}$.
Each value of the segment $\overline{OA}=a$ where $0\lt \overline{OA}\lt R\cos(\alpha)$ determines both the angle $\theta$ and the "little" radius $r$. You have $$\begin{cases}r=\sqrt{R^2+a^2-2aR\cos(\alpha)}\\\theta=\arcsin\left(\dfrac{R\sin(\alpha)}{r}\right)\end{cases}\qquad(*)$$
Finally your equation is $$2r\theta=R$$ that is
$$2\sqrt{R^2+a^2-2aR\cos(\alpha)}\arcsin\left(\dfrac{R\sin(\alpha)}{r}\right)=R\qquad(**)$$ You must put in $(**)$ the value of $r$ given in $(*)$, of course. You finally have an equation in one variable $a$ since $R$ and $\alpha$ are data of the problem.
Best Answer
Hint. Recall that the length of an arc of a circle and the corresponding angle subtended at the centre are proportional.