In triangle ABC, angle A = 40°, angle B = 70°, and side a = 12. Which value is closest to side b?

Use the Law of Sines, $\frac{a}{\sin A}=\frac{b}{\sin B}$. Then $b=\frac{12\sin70^\circ}{\sin40^\circ}\approx\frac{12(0.9397)}{0.6428}\approx17.5$, choice H. A sensible check is that the larger angle, $70^\circ$, lies opposite the longer side, so $b$ must be greater than $a=12$. That immediately eliminates choice F ($8.2$) and choice G ($12.0$). Choice F is what you get by inverting the ratio, $\frac{12\sin40^\circ}{\sin70^\circ}\approx8.2$. Choice J ($21.0$) is too large; it does not match the ratio $\frac{\sin70^\circ}{\sin40^\circ}\approx1.46$. The answer must exceed $12$ but only by a modest factor.