Ans: C
$\begin{array}{cl}
& \dfrac{30}{3\sin^2\theta+2\sin^2(90^\circ-\theta)} \\
= & \dfrac{30}{3\sin^2\theta+2\cos^2\theta} \\
= & \dfrac{30}{\sin^2\theta+2(\sin^2 \theta+\cos^2 \theta)} \\
= & \dfrac{30}{\sin^2\theta+2}
\end{array}$
$\begin{array}{cl}
& \dfrac{30}{3\sin^2\theta+2\sin^2(90^\circ-\theta)} \\
= & \dfrac{30}{3\sin^2\theta+2\cos^2\theta} \\
= & \dfrac{30}{\sin^2\theta+2(\sin^2 \theta+\cos^2 \theta)} \\
= & \dfrac{30}{\sin^2\theta+2}
\end{array}$
Hence, we have
$\begin{array}{rcl}
-1 \le & \sin \theta & \le 1 \\
0 \le & \sin^2\theta & \le 1 \\
2 \le & \sin^2\theta + 2 & \le 3 \\
\dfrac{1}{2} \ge & \dfrac{1}{\sin^2\theta+2} & \ge \dfrac{1}{3} \\
\dfrac{30}{2} \ge & \dfrac{30}{\sin^2\theta+2} & \ge \dfrac{30}{3} \\
15 \ge & \dfrac{1}{\sin^2\theta+2} & \ge 10 \\
\end{array}$
Therefore the least value is $10$.
