How do you determine the number of triangles using SSA?

Question

Suppose I'm given 2 sides and an angle for a triangle. How do I use those sides to determine whether the measurements can give 0, 1, or 2 triangles?

Do I use Law of Sines or Cosines?

Answer 1

You use neither the Law of Sines nor the Law of Cosines, but you may need to use the sine of the given angle.

You don't say in the main text of your question, but I'll assume the given angle is not the one between the two given sides, so it is indeed a SSA problem. Let's use the standard triangle $\triangle ABC$, where side $a$ is opposite vertex $A$ and angle $\alpha$ is at vertex $A$ (and similarly for the other sides and angles). Let's also say the given angle is $\alpha$ and the given sides are $a$ and $b$. First we look at obvious "degenerate" triangles.

1. If either of the sides is non-positive, there are $0$ triangles.
2. If it is not true that $0°<\alpha<180°$, there are $0$ triangles.

In the remaining cases, we'll assume that $a>0,b>0,0°<\alpha<180°$.

3. If $\alpha\ge 90°$ and $a\le b$, there are $0$ triangles.

4. If $\alpha\ge 90°$ and $a>b$, there is $1$ triangle.

5. If $\alpha<90°$ and $a<b\sin\alpha$, there are $0$ triangles.

6. If $\alpha<90°$ and $a=b\sin\alpha$, there is $1$ triangle, and $\beta=90°$.

7. If $\alpha<90°$ and $b\sin\alpha<a<b$, there are $2$ triangles, and the two possible values of $\beta$ are supplementary.

8. If $\alpha<90°$ and $b\le a$, there is $1$ triangle.

Proving these rigorously may be difficult, but these diagrams should make things clear.