Answer
two triangles.
$C_1\approx53.5^\circ$
$A_1\approx86.5^\circ$
$a_1\approx6.21$
$C_2\approx126.5^\circ$
$A_2\approx13.5^\circ$
$a_2\approx1.45$
Work Step by Step
1. Use the Law of Sines, we have $\frac{5}{sinC}=\frac{4}{sin40^\circ}=\frac{a}{sinA}$
2. $C_1=sin^{-1}(\frac{5sin40^\circ}{4})\approx53.5^\circ$ or $C_2\approx(180-57.7)=126.5^\circ$ two triangles.
3. Find the third angle $A_1\approx180-40-53.5=86.5^\circ$ or $A_2\approx180-40-126.5=13.5^\circ$
4. $a_1=\frac{4sin86.5^\circ}{sin40^\circ}\approx6.21$ or $a_2=\frac{4sin13.5^\circ}{sin40^\circ}\approx1.45$
