Question

When using the law of sines, why can the SSA case result in zero, one, or two triangles?

Answer 1

`sin B = b/a sin A` may be less than 1 giving a pair of supplementary angles for B; equal to 1, giving exactly `90^circ` for B, a single triangle; or greater than 1, giving no angle for B and no such triangle.

Explanation:

When we have two sides and an angle there are two cases: If the angle is the included angle, formed by the two sides, the triangle is uniquely determined and the path to solving starts with the Law Of Cosines.

The other case is the angle isn't the included angle, succinctly described as SSA. Say we have sides `a,b` and angle `A` opposite `a`. The Law of Sines let's us solve for `sin B`:

`a/sin A = b/sin B`

`sin B = b/a sin A`

Now we have the three cases mentioned in the question. First we note none of the quantities are negative; for triangle angles the sine is always positive.

The expression on the right may evaluate to more than 1. In that case, there's no such triangle that satisfies the givens.

Or the expression may evaluate to exactly 1, `sin A=1`, meaning A is `90^circ,` so the problem determines exactly one triangle, a right triangle.

Finally, the most common case is `sin A<1.` There are in general two triangles angles with a given sine. The principal angle will be in the first quadrant, and the other angle is the supplement (the angles add to `180^circ`), in the second quadrant.

The two possible angles for A give two possible triangles ABC which satisfy the given SSA.

We might want to single out `sin B=0` as a special case. That's called the degenerate triangle, zero area. It's not really a triangle, more like a line segment. The Law of Sines really only applies to real triangles; we get division by zero in these cases.