You're correct to look for right-angled triangles. In particular, as shown below in a modified version of your diagram, the midpoint of $AB$ is labeled as $C$, and with the line segments of $SA$, $SC$ and $SB$ added.

As Divide1918's comment indicates, $\lvert SA\rvert = \lvert SB\rvert = 5$ as they are the radii. Thus, $\triangle ASB$ is isosceles, so with $C$ being the midpoint of $AB$, then $\measuredangle SCB = 90^{\circ}$. Since $\lvert CB\rvert = 4$, then using the Pythagorean theorem with $\triangle SBC$, we get that
$$\lvert SC\rvert = \sqrt{5^2 - 4^2} = \sqrt{9} = 3$$
With $\triangle SCR$ being right-angled, we have
$$\sin(x) = \frac{\lvert SC\rvert}{\lvert SR\rvert} = \frac{3}{12} = 0.25 \;\;\;\to\;\;\; x = \sin^{-1}(0.25)$$