Answer
two triangles.
$C_1\approx74.6^\circ$
$A_1\approx65.4^\circ$
$a_1\approx2.83$
$C_2\approx105.4^\circ$
$A_2\approx34.6^\circ$
$a_2\approx1.77$
Work Step by Step
1. Use the Law of Sines, we have $\frac{3}{sinC}=\frac{2}{sin40^\circ}=\frac{a}{sinA}$
2. $C_1=sin^{-1}(\frac{3sin40^\circ}{2})\approx74.6^\circ$ or $C_2\approx(180-74.6)^\circ=105.4^\circ$ two triangles.
3. Find the third angle $A_1\approx180-40-74.6=65.4^\circ$ or $A_2\approx180-40-105.4=34.6^\circ$
4. $a_1=\frac{2sin65.4^\circ}{sin40^\circ}\approx2.83$ or $a_2=\frac{2sin34.6^\circ}{sin40^\circ}\approx1.77$
