Answer
$B_1 \approx13.4^\circ$
$C_1\approx 156.6^\circ$
$c_1 \approx6.86$
$B_2\approx166.6^\circ$
$C_2\approx 3.4^\circ$
$c_2 \approx1.02$
Work Step by Step
1. Use the Law of Sines $\frac{3}{sin10^\circ}=\frac{4}{sinB}=\frac{c}{sinC}$
2. Thus $B_1=sin^{-1}(\frac{4sin10^\circ}{3})\approx13.4^\circ$ or $B_2\approx166.6^\circ$
3. Find the third angle $C_1\approx180-10-13.4=156.6^\circ$ or $C_2\approx180-10-166.6=3.4^\circ$
4. $c_1=\frac{3sin156.6^\circ}{sin10^\circ}\approx6.86$ or
$c_2=\frac{3sin3.4^\circ}{sin10^\circ}\approx1.02$
