The law of sines, $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$, solves a triangle when you know two angles and a side. Something odd happens when you know two sides and an angle that is not between them (side-side-angle, SSA): the measurements might fit no triangle, one, or two.
Key idea
Given angle $A$, the side $a$ across from it, and another side $b$, compute $\sin B = \dfrac{b\sin A}{a}$. If $\sin B > 1$, no triangle. Otherwise both $B$ and $180^\circ - B$ are candidates; keep each one with $A + B < 180^\circ$. For an acute $A$ this means, with the height $h = b\sin A$: $a < h$ none; $a = h$ one right triangle; $h < a < b$ two; $a \ge b$ one.
Why it works
Fix angle $A$ and side $b = AC$. Side $a$ hangs from $C$ like the arm of a compass, and $B$ is where it lands on the other ray. If $a$ is shorter than the height $h$, it cannot reach. If it is exactly $h$, it just touches. If it is a little longer, it can swing to land on either side of the foot of the height: two triangles. In the algebra, this shows up because sine is positive in Quadrants I and II, so two angles have the same sine, and $\sin^{-1}$ only reports the acute one.
Worked examples
Worked example: Two triangles
In $\triangle ABC$, $A = 30^\circ$, $a = 4$ and $b = 4\sqrt{2}$. Find every possible $B$.
- $\sin B = \dfrac{b\sin A}{a} = \dfrac{4\sqrt{2} \cdot \frac{1}{2}}{4} = \dfrac{\sqrt{2}}{2}$.
- Candidates: $B = 45^\circ$ or $B = 180^\circ - 45^\circ = 135^\circ$.
- Check: $30^\circ + 45^\circ < 180^\circ$ and $30^\circ + 135^\circ = 165^\circ < 180^\circ$. Both leave room for $C$.
Answer: $B = 45^\circ$ (then $C = 105^\circ$) or $B = 135^\circ$ (then $C = 15^\circ$)
Worked example: Counting with the height
$A = 30^\circ$, $b = 10$, $a = 6$. How many triangles?
- $h = b\sin A = 10 \cdot \frac{1}{2} = 5$.
- $5 < 6 < 10$, so $h < a < b$: side $a$ can land in two places.
- (With a calculator: $\sin B = \frac{5}{6}$, so $B \approx 56.4^\circ$ or $B \approx 123.6^\circ$.)
Answer: Two triangles
Worked example: An obtuse angle
$A = 120^\circ$, $a = 5$, $b = 8$. How many triangles?
- An obtuse angle must be the largest angle, so the side across from it, $a$, must be the longest side.
- Here $a = 5 < b = 8$.
Answer: No triangle
Common mistake
Finding one angle with $\sin^{-1}$ and stopping. The calculator gives only the acute angle; the obtuse angle $180^\circ - B$ has the same sine and often works too. Always test the second candidate: does $A + (180^\circ - B)$ stay under $180^\circ$?