Answer
Triangle 1: $A \approx59.0^\circ$, $B \approx81.0^\circ$, $b \approx23.05$;
Triangle 2: $A \approx121.0^\circ$, $B \approx19.0^\circ$, $b \approx7.59$
Work Step by Step
Step 1. Given $a=20, c=15, C=40^\circ$, use the Law of Sines, we have $\frac{sinA}{20}=\frac{sin40^\circ}{15}$, thus $A=sin^{-1}(\frac{20sin40^\circ}{15})\approx59.0^\circ$ or $A\approx121.0^\circ$;
Step 2. For $A \approx59.0^\circ$, we have $B=180-A-C\approx81.0^\circ$ and $\frac{sin81.0^\circ}{b}=\frac{sin40^\circ}{15}$, thus $b=\frac{15sin81.0^\circ}{sin40^\circ}\approx23.05$;
Step 3. For $A \approx121.0^\circ$, we have $B=180-A-C\approx19.0^\circ$ and $\frac{sin19.0^\circ}{b}=\frac{sin40^\circ}{15}$, thus $b=\frac{15sin19.0^\circ}{sin40^\circ}\approx7.59$.
