In the last few posts, we studied the SSA case of solving for a triangle, when two sides and an non-included angle are given. (Some mathematics instructors happily prefer the angle-side-side acronym to bluntly describe the complications that arise from this possibly ambiguous case. I personally prefer not to use this acronym.)

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:https://meangreenmath.files.wordpress.com/2014/10/philistines.png

Suppose that , , and the nonincluded angle are given, and we are supposed to solve for , , and . As we’ve seen in this series, there are four distinct cases — and handling these cases requires accurately solving equation like on the interval .

**Case 1**. . In this case, there are **no solutions**. When the Law of Sines is employed and we reach the step

the is greater than 1, which is impossible.

**Case 2**. . This rarely arises in practice (except by careful writers of textbooks). In this case, there is exactly **one solution**. When the Law of Sines is employed, we obtain

We conclude that , so that is a right triangle.

**Case 3**. . This is the ambiguous case that yields **two solutions**. The Law of Sines yields

so that there are two possible choices for , and .

**Case 4**. . This yields **one solution**. Similar to Case 3, the Law of Sines yields

so that there are two possible choices for , and . However, when the second larger value of is attempted, we end up with a negative angle for , which is impossible (unlike Case 3).

Many mathematics students prefer to memorize rules like those listed above. However, I try to encourage my students not to blindly use rules when solving the SSA case, as it’s just too easy to make a mistake in identifying the proper case. Instead, I encourage them to use the Law of Sines and to remember that the equation

has two solutions in as long as $0 < t < 1$:

If they can remember this fact, then students can just follow their noses when applying the Law of Sines, identifying impossible and ambiguous cases when the occasions arise.

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