Answer
two triangles.
$A_1 \approx59.0^\circ$
$B_1\approx 81.0^\circ$
$b_1 \approx23.05$
$A_2\approx121.0^\circ$
$B_2\approx 19.0^\circ$
$b_2 \approx7.59$
Work Step by Step
1. Use the Law of Sines $\frac{20}{sinA}=\frac{b}{sinB}=\frac{15}{sin40^\circ}$
2. Thus $A_1=sin^{-1}(\frac{20sin40^\circ}{15})\approx59.0^\circ$ or $A_2\approx121.0^\circ$, two triangles.
3. Find the third angle $B_1\approx180-40-59.0=81.0^\circ$ or $B_2\approx180-40-121.0=19.0^\circ$
4. $b_1=\frac{15sin81.0^\circ}{sin40^\circ}\approx23.05$ or
$b_2=\frac{15sin19.0^\circ}{sin40^\circ}\approx7.59$
