Answer
two triangles.
$A_1\approx57.7^\circ$
$B_1\approx97.3^\circ$
$b_1\approx2.35$
$A_2\approx122.3^\circ$
$B_2\approx32.7^\circ$
$b_2\approx1.28$
Work Step by Step
1. Use the Law of Sines, we have $\frac{2}{sinA}=\frac{1}{sin25^\circ}=\frac{b}{sinB}$
2. $A_1=sin^{-1}(\frac{2sin25^\circ}{1})\approx57.7^\circ$ or $A_2\approx(180-57.7)=122.3^\circ$ two triangles.
3. Find the third angle $B_1\approx180-25-57.7=97.3^\circ$ or $B_2\approx180-25-122.3=32.7^\circ$
4. $b_1=\frac{sin97.3^\circ}{sin25^\circ}\approx2.35$ or $b_2=\frac{sin32.7^\circ}{sin25^\circ}\approx1.28$
