# How do you determine whether triangle ABC has no, one, or two solutions given A=30^circ, a=3, b=6?

In this particular case we're given $A = {30}^{\circ}$ and $a = 3$, which is the side opposite $A$. Since $b = 6 = 2 a$ we actually know that we're dealing with a ${30}^{\circ}$-${60}^{\circ}$-${90}^{\circ}$ triangle because of the ratio of sides: $x - x \sqrt{3} - 2 x$. In this case there is exactly one triangle and we don't really need to use the Law of Sines.
If you want to use the law of sines, though, calculate $6 \cdot \sin \left({30}^{\circ}\right) = 6 \left(\frac{1}{2}\right) = 3$ which is exactly the length of $a$, the side opposite the given angle, which means we have one right triangle solution.