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 exactly one 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:
Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.
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:
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 yesterday’s post, there are two angles between and with a sine of :
or, in degrees,
So we have two different cases to check.
Case 1: . We begin by solving for :
Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find . In this case, it’s best to use the pair instead of since the values of and are both known exactly.
Case 2: . We again begin by solving for :
Oops. That’s clearly impossible. So there is only one possible triangle, and the missing pieces are , , and . Judging from the above (correctly drawn) picture, these numbers certainly look plausible.
It turns out that Case 2 will always fail in SSA will always fail as long as the side opposite the given angle is longer than the other given side (in this case, ). However, I prefer that my students not memorize this rule. Instead, I’d prefer that they list the two possible values of and then run through the logical consequences, stopping when an impossibility is reached. As we’ll see in tomorrow’s post, it’s perfectly possible for Case 2 to produce a second valid solution with the proper choice for the length of the side opposite the given angle.