Step 1: Use exact trigonometric values instead of decimals.
For $\theta = 30^\circ$, the exact values are $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$ and $\sin 30^\circ = \dfrac{1}{2}$. Substitute these into the rotation acting on $(5, 3)$.
Step 2: Compute the rotated vector symbolically.
\[ x_r = 5\cos30^\circ - 3\sin30^\circ = \frac{5\sqrt3 - 3}{2},\qquad y_r = 5\sin30^\circ + 3\cos30^\circ = \frac{5 + 3\sqrt3}{2} \]Using $\sqrt3 \approx 1.7321$: $x_r = \dfrac{8.6603 - 3}{2} = 2.8301$ and $y_r = \dfrac{5 + 5.1962}{2} = 5.0981$.
Step 3: Scale by $s = 2$ and add the translation $[1, 2]$.
\[ x' = 2(2.8301) + 1 = 6.6603 \approx 6.66,\qquad y' = 2(5.0981) + 2 = 12.1962 \approx 12.20 \]Both components round exactly to the target vector, confirming $s = 2$, $\theta = 30^\circ$, $[t_x,t_y] = [1,2]$.
Step 4: Rule out the other candidates on physical grounds.
The output vector $(6.66, 12.20)$ has magnitude about $13.9$, noticeably larger than the original vector's magnitude $\sqrt{34} \approx 5.83$, so the scale factor must be greater than 1, which rules out options (B) and (D), whose scale factors ($1/2$ and $1/\sqrt2$) would shrink the rotated vector before translation is added. Between the remaining candidates with $s>1$, option (A) with $\theta=30^\circ$ and option (C) with $\theta=-30^\circ$, only $\theta = 30^\circ$ gives the exact match computed above.\[ \boxed{s = 2,\ \theta = 30^\circ,\ [t_x, t_y] = [1, 2]} \]