# Inverse functions: Arcsine and SSA (Part 17)

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 $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:https://meangreenmath.files.wordpress.com/2014/10/philistines.png Suppose that $a$, $c$, and the nonincluded angle $\alpha$ are given, and we are supposed to solve for $b$, $\beta$, and $\gamma$. As we’ve seen in this series, there are four distinct cases — and handling these cases requires accurately solving equation like $\sin \gamma = \hbox{something}$ on the interval $[0^\circ, 180^\circ]$.

Case 1. $b < c \sin \alpha$. In this case, there are no solutions. When the Law of Sines is employed and we reach the step $\sin \gamma = \hbox{something}$

the $\hbox{something}$ is greater than 1, which is impossible. Case 2. $b = c \sin \alpha$. 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 $\sin \gamma = 1$

We conclude that $\gamma = 90^\circ$, so that $\triangle ABC$ is a right triangle. Case 3. $c \sin \alpha < b < c$. This is the ambiguous case that yields two solutions. The Law of Sines yields $\sin \gamma = \hbox{something}$

so that there are two possible choices for $\gamma$, $\hbox{some angle}$ and $180^\circ - \hbox{some angle}$. Case 4. $b > c$. This yields one solution. Similar to Case 3, the Law of Sines yields $\sin \gamma = \hbox{something}$

so that there are two possible choices for $\gamma$, $\hbox{some angle}$ and $180^\circ - \hbox{some angle}$. However, when the second larger value of $\gamma$ is attempted, we end up with a negative angle for $\beta$, 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 $\sin \gamma = t$

has two solutions in $[0^\circ, 180^\circ]$ as long as $0 < t < 1$: $\gamma = \sin^{-1} t \qquad \hbox{and} \qquad \gamma = 180^\circ - \sin^{-1} t$

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.

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.