$\sin 2\alpha = \dfrac{\sqrt{3}}{2} \Rightarrow 2\alpha = 60° \Rightarrow \alpha = 30°$. Therefore, $\sin 3\alpha = \sin 90° = \mathbf{1}$. Answer: (C)
From Table 8.1, $\sin 60° = \dfrac{\sqrt{3}}{2}$, so $2\alpha = 60°$, giving $\alpha = 30°$. Then $3\alpha = 90°$ and $\sin 90° = 1$. Students must recall standard angle values from the table and work step-by-step.