Answer
two triangles.
$B_1 \approx35.4^\circ$
$C_1\approx134.6^\circ$
$c_1 \approx12.29$,
$B_2\approx144.6^\circ$,
$C_2\approx25.4^\circ$
$c_2 \approx7.40$
Work Step by Step
1. Use the Law of Sines, we have $\frac{3}{sin10^\circ}=\frac{10}{sinB}=\frac{c}{sinC}$, thus $B_1=sin^{-1}(\frac{10sin10^\circ}{3})\approx35.4^\circ$ or $B_2\approx144.6^\circ$, two triangles.
2. Find the third angle $C_1\approx(180-10-35.4)=134.6^\circ$ or $C_2\approx(180-10-144.6)=25.4^\circ$
3. thus $c_1=\frac{3sin134.6^\circ}{sin10^\circ}\approx12.29$, $c_2=\frac{3sin25.4^\circ}{sin10^\circ}\approx7.40$
