# Inverse Functions: Arctangent and Angle Between Two Lines (Part 24)

Here’s a straightforward application of arctangent that, a generation ago, used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry).

Find the smallest angle between the lines $y= 3x$ and $y = -x/2$.

This problem is almost equivalent to finding the angle between the vectors $\langle 1,3 \rangle$ and $\langle -2,1 \rangle$. I use the caveat almost because the angle between two vectors could be between $0$ and $\pi$, while the smallest angle between two lines must lie between $0$ and $\pi/2$.

This smallest angle can be found using the formula

$\theta = \displaystyle \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$,

where $m_1$ and $m_2$ are the slopes of the two lines. In the present case,

$\theta = \tan^{-1} \left( \left| \displaystyle \frac{ 3 - (-1/2) }{1 + (3)(-1/2)} \right| \right)$

$\theta = \tan^{-1} \left( \left| \displaystyle \frac{7/2}{-1/2} \right| \right)$

$\theta = \tan^{-1} 7$

$\theta \approx 81.87^\circ$.

Not surprisingly, we obtain the same answer that we obtained a couple of posts ago using arccosine. The following picture makes clear why $\tan^{-1} 7 = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}$.

In tomorrow’s post, I’ll explain why the above formula actually works.

# Inverse Functions: Arccosine and Dot Products (Part 23)

The Law of Cosines can be applied to find the angle between two vectors ${\bf a}$ and ${\bf b}$. To begin, we draw the vectors ${\bf a}$ and ${\bf b}$, as well as the vector ${\bf c}$ (to be determined momentarily) that connects the tips of the vectors ${\bf a}$ and ${\bf b}$.

Using the usual rules for adding vectors, we see that ${\bf a} + {\bf c} = {\bf b}$, so that ${\bf c} = {\bf b} - {\bf a}$

We now apply the Law of Cosines to find $\theta$:

$\parallel \! \! {\bf c} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

$\parallel \! \! {\bf b} - {\bf a} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

We now apply the rule $\parallel \! \! {\bf a} \! \! \parallel^2 = {\bf a} \cdot {\bf a}$, convert the square of the norms into dot products. We then use the distributive and commutative properties of dot products to simplify.

$( {\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot ({\bf b} - {\bf a}) - {\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot ({\bf b} - {\bf a}) -{\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot {\bf b} - {\bf a} \cdot {\bf b} - {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot {\bf b} - 2 {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

We can now cancel from the left and right sides and solve for $\cos \theta$:

$- 2 {\bf a} \cdot {\bf b} = - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

$\displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } = \cos \theta$

Finally, we are guaranteed that the angle between two vectors must lie between $0$ and $\pi$ (or, in degrees, between $0^\circ$ and $180^\circ$). Since this is the range of arccosine, we are permitted to use this inverse function to solve for $\theta$:

$\cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } \right) = \theta$

The good news is that there’s nothing special about two dimensions in the above proof, and so this formula may used for vectors in $\mathbb{R}^n$ for any dimension $n \ge 2$.

In the next post, we’ll consider how this same problem can be solved — but only in two dimensions — using arctangent.

# Inverse Functions: Arccosine and Dot Products (Part 22)

Here’s a straightforward application of arccosine, that, as far as I can tell, isn’t taught too often in Precalculus and is not part of the Common Core standards for vectors and matrices.

Find the angle between the vectors $\langle 1,3 \rangle$ and $\langle -2,1 \rangle$.

This problem is equivalent to finding the angle between the lines $y = 3x$ and $y = -x/2$. The angle $\theta$ is not drawn in standard position, which makes measurement of the angle initial daunting.

Fortunately, there is the straightforward formula for the angle between two vectors ${\bf a}$ and ${\bf b}$:

$\theta = \cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \!\! {\bf a} \!\! \parallel \parallel \!\! {\bf b} \!\! \parallel } \right)$

We recall that ${\bf a} \cdot {\bf b}$ is the dot product (or inner product) of the two vectors ${\bf a}$ and ${\bf b}$, while $\parallel \!\! {\bf a} \!\! \parallel = \sqrt{ {\bf a} \cdot {\bf a} }$ is the norm (or length) of the vector ${\bf a}$.

For this particular example,

$\theta = \cos^{-1} \left( \displaystyle \frac{\langle 1,3 \rangle \cdot \langle -2,1 \rangle }{ \parallel \!\!\langle 1,3 \rangle \!\! \parallel \parallel \!\!\langle -2,1 \rangle \!\! \parallel } \right)$

$\theta = \cos^{-1} \left( \displaystyle \frac{ (1)(-2) + (3)(1) }{ \sqrt{ (1)^2 + (3)^2} \sqrt{ (-2)^2 + 1^2 }} \right)$

$\theta = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}$

$\theta \approx 81.87^\circ$

In the next post, we’ll discuss why this actually works. And then we’ll consider how the same problem can be solved more directly using arctangent.

# Inverse Functions: Arccosine and SSS (Part 21)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC$ if $a = 16$, $b = 20$, and $c = 25$.

When solving for the three angles, it’s best to start with the biggest angle (that is, the angle opposite the biggest side). To see why, let’s see what happens if we first use the Law of Cosines to solve for one of the two smaller angles, say $\alpha$:

$a^2 = b^2 + c^2 - 2 b c \cos \alpha$

$256 = 400 + 625 - 1000 \cos \alpha$

$-769 = -1000 \cos \alpha$

$0.769 = \cos \alpha$

$\alpha \approx 39.746^\circ$

So far, so good. Now let’s try using the Law of Sines to solve for $\gamma$:

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

$\displaystyle \frac{\sin 39.746^\circ}{16} \approx \displaystyle \frac{\sin \gamma}{25}$

$0.99883 \approx \sin \gamma$

Uh oh… there are two possible solutions for $\gamma$ since, hypothetically, $\gamma$ could be in either the first or second quadrant! So we have no way of knowing, using only the Law of Sines, whether $\gamma \approx 87.223^\circ$ or if $\gamma \approx 180^\circ - 87.223^\circ = 92.777^\circ$.

For this reason, it would have been far better to solve for the biggest angle first. For the present example, the biggest answer is $\gamma$ since that’s the angle opposite the longest side.

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.223^\circ$.

We now use the Law of Sines to solve for either $\alpha$ or $\beta$ (pretending that we didn’t do the work above). Let’s solve for $\alpha$:

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

$\displaystyle \frac{\sin \alpha}{16} \approx \displaystyle \frac{\sin 87.223}{25}$

$\sin \alpha \approx 0.63949$

This equation also has two solutions in the interval $[0^\circ, 180^\circ]$, namely, $\alpha \approx 39.736^\circ$ and $\alpha \approx 180^\circ - 39.736^\circ = 140.264^\circ$. However, we know full well that the answer can’t be larger than $\gamma$ since that’s already known to be the largest angle. So there’s no need to overthink the matter — the answer from blindly using arcsine on a calculator is going to be the answer for $\alpha$.

Naturally, the easiest way of finding $\beta$ is by computing $180^\circ - \alpha - \gamma$.

# Inverse Functions: Arccosine and SSS (Part 20)

The Law of Cosines also recognizes when the purported sides of a triangle are impossible.

Solve $\triangle ABC if$latex a = 16$, $b = 20$, and $c = 40$. Hopefully students would recognize that $c > a + b$, thus quickly demonstrating that the triangle is impossible. However, this also falls out of the Law of Cosines: $c^2 = a^2 + b^2 - 2 a b \cos \gamma$ $1600 = 256 + 400 - 640 \cos \gamma$ $944 =-640 \cos \gamma$ $-1.475 = \cos \gamma$ Since the cosine of an angle can’t be less than -1, we can conclude that this is impossible. Stated another way, we have the implications (since $a$, $b$, and $c$ are all positive) $c > a + b \Longleftrightarrow c^2 > (a+b)^2$ $\Longleftrightarrow a^2 + b^2 - 2 a b \cos \gamma > a^2 + 2 a b + b^2$ $\Longleftrightarrow -2 a b \cos \gamma > 2 a b$ $\Longleftrightarrow \cos \gamma < -1$ Since the last statement is impossible, so is the first one. # Inverse Functions: Arccosine and SSS (Part 19) Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle. Solve $\triangle ABC if$latex a = 16$, $b = 20$, and $c = 25$.

To solve for, say, the angle $\gamma$, we employ the Law of Cosines:

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.2^\circ$. And the good news is that there is no need to overthink this… this is guaranteed to be the angle since the range of $y = \cos^{-1} x$ is $[0,\pi]$, or $[0^\circ, 180^\circ]$ in degrees. So the equation

$\cos x = \hbox{something}$

is guaranteed to have a unique solution between $0^\circ$ and $180^\circ$. (But there are infinitely many solutions on $\mathbb{R}$. And since an angle in a triangle must lie between $0^\circ$ and $180^\circ$, the practical upshot is that just plugging into a calculator blindly is perfectly OK for this problem. This is in stark contrast to the Law of Sines, for which some attention must be paid for solutions in the interval $[0^\circ,90^\circ]$ and also the interval $[90^\circ, 180^\circ]$.

From this point forward, the Law of Cosines could be employed again to find either $\alpha$ or $\beta$. Indeed, this would be my preference since the sides $a$, $b$, and $c$ are exactly. However, my experience is that students prefer the simplicity of the Law of Sines to solve for one of these angles, using the now known pair of $c$ (exactly known) and $\gamma$ (approximately known with a calculator).

# Inverse Functions: Arccosine and Arctangent (Part 18)

In this series, we’ve seen that the inverse of function that fails the horizontal line test can be defined by appropriately restricting the domain of the function. For example, we now look at the graph of $y = \cos x$:

As with the graph of $y = \sin x$, we select a section that satisfies the horizontal line test and ignore the rest of the graph. (I described this in more rigorous terms when I considered arcsine, so I will not repeat the rigor here.) There are plenty of choices that could be made; by tradition, the interval $[0,\pi]$ is chosen.

Reflecting only the half-wave of the cosine graph on the interval $[0,\pi]$ through the line $y = x$ produces the graph of $y= \cos^{-1} x$. Again, to assist my students when graphing this function, I point out that the graph of cosine has horizontal tangent lines at the points $(0,1)$ and $(\pi,-1)$. Therefore, after reflecting through the line $y = x$, we see that the graph of $\cos^{-1} x$ has vertical tangent lines at $(1,0)$ and $(-1,\pi)$.

The same logic applies when defining the arctangent function. By tradition, the interval $(-\pi/2,\pi/2)$ is chosen as the section of the graph of $y = \tan x$ that satisfies the horizontal line test.

Reflecting only the half-wave of the cosine graph on the interval $[0,\pi]$ through the line $y = x$ produces the graph of $y= \tan^{-1} x$. Like the (more complicated) logistic growth function, this function has two different horizontal asymptotes that govern the behavior of the function as $x \to \pm \infty$.

So here are the rules that I want my Precalculus students to memorize:

$y = \sin^{-1} x$ means that $x = \sin y$ and $\displaystyle -\frac{\pi}{2} \le y \le \displaystyle \frac{\pi}{2}$

$y = \cos^{-1} x$ means that $x = \cos y$ and $0 \le y \le \pi$

$y = \sin^{-1} x$ means that $x = \sin y$ and $\displaystyle -\frac{\pi}{2} < y < \displaystyle \frac{\pi}{2}$

Students using forget that the range of arccosine is different than the other two, and I’ll usually have to produce the graph of $y = \cos x$ to explain and re-explain why this one is different.

Because these functions are defined on restricted domains, the usual funny things can happen. For example,

$\cos^{-1} (\cos 2\pi) = \cos^{-1} 1 = 0 \ne 2 \pi$

$\tan^{-1} \left( \tan \displaystyle \frac{3\pi}{4} \right) = \tan^{-1} (-1) = -\displaystyle \frac{\pi}{4} \ne \displaystyle \frac{3\pi}{4}$

# 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.

# 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:

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:

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

This 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 $\sin^{-1} \frac{1}{3}$.

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.

# 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:

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:

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. 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 $\sin^{-1} \frac{1}{3}$.

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.