Inverse functions: Arcsine and SSA (Part 15)

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 \sin \theta = 0.8. 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 \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:

philistines

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 \triangle ABC if a = 15, c = 10, and \alpha = 30^\circ.

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

badSSA3

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

SSA3

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:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1}{3} = \sin \gamma

At this point, the natural inclination of a student is to pop out the calculator and find \sin^{-1} \frac{1}{3}.

SSAcalc1

This is incorrect logic that, as discussed extensively in yesterday’s post, there are two angles between 0^\circ and 180^\circ with a sine of 1/3:

\sin^{-1} \frac{1}{3} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{1}{3},

or, in degrees,

\gamma \approx 19.47^\circ \qquad \hbox{and} \qquad \gamma \approx 160.53^\circ

So we have two different cases to check.

Case 1: \gamma \approx 19.47^\circ. We begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 130.53^\circ

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find b. In this case, it’s best to use the pair \alpha - a instead of \gamma - c since the values of \alpha and a are both known exactly.

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}

\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin 130.53^\circ}{b}

b = \displaystyle \frac{15 \sin 130.53^\circ}{\sin 30^\circ}

b \approx 22.8

Case 2: \gamma \approx 130.53^\circ. We again begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx -10.53^\circ

Oops. That’s clearly impossible. So there is only one possible triangle, and the missing pieces are \gamma \approx 19.47^\circ, \beta \approx 130.53^\circ, and b \approx 22.8. 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, a > c). However, I prefer that my students not memorize this rule. Instead, I’d prefer that they list the two possible values of \gamma 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.

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