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 **no** solution:

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:

The red dashed circle with center illustrates the dilemma: “side” is simply too short to reach the horizontal dashed line to make the vertex , dangling limply from the vertex .

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

Since must like between and (said another way, is undefined), we know that this triangle cannot be solved.

In the next few posts of this series, I’ll consider the other SSA cases — including the case where *two* solutions are possible.

### Like this:

Like Loading...

*Related*