# Inverse functions: Arcsine and SSA (Part 13)

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 exactly one solution:

Solve $\triangle ABC$ if $a = 5$, $c = 10$, 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:

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.

Of course, the reason that the dashed circle and line intersect at exactly one point is because $a = c \sin \alpha$, so that the triangle is a right triangle.

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:

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

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle 1 = \sin \gamma$

$90^\circ = \gamma$

The jump to the last step is only possible because there’s exactly one angle between $0^\circ$ and $90^\circ$ whose sine is equal to $1$. In the next couple posts in this series, we’ll see what happens when we get a step where $0 < \sin \gamma < 1$.

Anyway, for the problem at hand, from this point forward it’s easy to solve for the remaining pieces. We begin by finding $\beta$:

$\beta = 180^\circ - \alpha - \gamma = 60^\circ$

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find $b$:

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

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 60^\circ}{b}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sqrt{3}/2}{b}$

$b = 5\sqrt{3}$

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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