We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like . In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of , will be the length of the side opposite , will be the length of the side opposite , and will be the length of the side opposite angle . Also will be the measure of , will be measure of , and will be the measure of . Modern textbooks tend not to use , , and for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has **two different **solutions:

Solve if , , and .

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

This time, the red circle intersects the dashed black line at **two different points**. So there will be **two different solutions** for this case. In other words, the phrasing of the question is somewhat deceptive. Usually when the question asks, “Solve the triangle…”, it’s presumed that there is only one solution. In this case, however, there are two different solutions.

These two different solutions appear when using the Law of Sines:

At this point, the natural inclination of a student is to pop out the calculator and find .

This is incorrect logic that, as discussed extensively in earlier in this series of posts, there are two angles between and with a sine of :

,

or, in degrees,

So we have two different cases to check. Unlike the previous posts in this series, it’s really, really important that we list both of these cases.

**Case 1: **. We begin by solving for :

Then we can use the Law of Sines to find . In this case, it’s best to use the pair instead of since the values of and are both known exactly.

This triangle with , , and corresponds to the bigger of the two triangles in the above picture, or the rightmost of the two places where the dotted circle intersects the black dotted line.

**Case 2: **. We again begin by solving for :

Unlike yesterday’s example, this is possible. So we have to continue the calculation to find :

This second triangle with , , and corresponds to the thinner of the two triangles in the above picture, or the leftmost of the two places where the dotted circle intersects the black dotted line.