# Inverse functions: Arcsine and SSA (Part 12)

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:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has no solution:

Solve $\triangle ABC$ if $a = 3$, $c = 10$, and $\alpha = 30^\circ$.

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

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

The red dashed circle with center $B$ illustrates the dilemma: “side” $BC$ is simply too short to reach the horizontal dashed line to make the vertex $C$, dangling limply from the vertex $B$.

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:

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

$\displaystyle \frac{\sin 30^\circ}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{5}{3} = \sin \gamma$

Since $\sin \gamma$ must like between $0$ and $1$ (said another way, $\sin^{-1} \frac{5}{3}$ is undefined), we know that this triangle cannot be solved.

In the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.

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