Inverse functions: Arcsine and SSA (Part 16)

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 two different solutions:

Solve \triangle ABC if a = 8, 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:

badSSA4

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

SSA4This time, the red circle intersects the dashed black line at two different points. So there will be two different solutions for this case. In other words, the phrasing of the question is somewhat deceptive. Usually when the question asks, “Solve the triangle…”, it’s presumed that there is only one solution. In this case, however, there are two different solutions.

These two different solutions appear when using the Law of Sines:

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

\displaystyle \frac{\sin 30^\circ}{8} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{8} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{5}{8} = \sin \gamma

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

SSAcalc4

This is incorrect logic that, as discussed extensively in earlier in this series of posts, there are two angles between 0^\circ and 180^\circ with a sine of 5/8:

\sin^{-1} \frac{5}{8} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{5}{8},

or, in degrees,

\gamma \approx 38.68^\circ \qquad \hbox{and} \qquad \gamma \approx 141.32^\circ

So we have two different cases to check. Unlike the previous posts in this series, it’s really, really important that we list both of these cases.

Case 1: \gamma \approx 38.68^\circ. We begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 111.32^\circ

Then we can use the Law of Sines 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}{5} = \displaystyle \frac{\sin 111.32^\circ}{b}

b = \displaystyle \frac{8 \sin 111.32^\circ}{\sin 30^\circ}

b \approx 14.9

This triangle with \gamma \approx 38.68^\circ, \beta \approx 111.32^\circ, and b \approx 14.9 corresponds to the bigger of the two triangles in the above picture, or the rightmost of the two places where the dotted circle intersects the black dotted line.

Case 2: \gamma \approx 141.32^\circ. We again begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 8.68^\circ

Unlike yesterday’s example, this is possible. So we have to continue the calculation to find b:

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

\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 8.68^\circ}{b}

b = \displaystyle \frac{8 \sin 8.68^\circ}{\sin 30^\circ}

b \approx 2.4

This second triangle with \gamma \approx 141.32^\circ, \beta \approx 8.68^\circ, and b \approx 2.4 corresponds to the thinner of the two triangles in the above picture, or the leftmost of the two places where the dotted circle intersects the black dotted line.

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