Langley’s Adventitious Angles: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my short series on a couple of easily stated but remarkably difficult geometry problems.

Part 1: The world’s second hardest easy geometry problem.

Part 2: The world’s hardest easy geometry problem.

Langley’s Adventitious Angles (Part 1)

Math With Bad Drawings had an interesting post about solving for $x$ in the following picture (this picture is taken from http://thinkzone.wlonk.com/MathFun/Triangle.htm):

I had never heard of this problem before, but it’s apparently well known and is called Langley’s Adventitious Angles. See Math With Bad Drawings, Wikipedia, and Math Pages for more information about the solution of this problem. Math Pages has a nice discussion about mathematical aspects of this problem, including connections to the Laws of Sines and Cosines and to various trig identities.

I’d encourage you to try to solve for $x$ without clicking on any of these links… a certain trick out of the patented Bag of Tricks is required to solve this problem using only geometry (as opposed to the Law of Cosines and the Law of Sines). I have a story that I tell my students about the patented Bag of Tricks: Socrates gave the Bag of Tricks to Plato, Plato gave it to Aristotle, it passed down the generations, my teacher taught the Bag of Tricks to me, and I teach it to my students. In the same post, Math With Bad Drawings has a nice discussion about pedagogical aspects of this problem concerning when a “trick” becomes a “technique”.

I recommend this problem for advanced geometry students who need to be challenged; even bright students will be stumped concerning coming up with the requisite trick on their own. Indeed, the problem still remains quite challenging even after the trick is shown.

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 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: 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).

SSA3 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.

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)

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.

Inverse functions: Arcsine and SSA (Part 14)

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:

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 we’ll see tomorrow, nevertheless leads to the correct conclusion. This is incorrect logic because there are two angles between $0^\circ$ and $180^\circ$ with a sine of $1/3$ . There is one solution in the first quadrant (the unique answer specified by arcsine), and there is another answer in the second quadrant — which is between $90^\circ$ and $180^\circ$ and hence not a permissible value of arcsine. Let me demonstrate this in three different ways.

First, let’s look at the graph of $y = \sin x$ (where, for convenience, the units of the $x-$ axis are in degrees). This graph intersects the line $y = \frac{1}{3}$ in two different places between $0^\circ$ and $180^\circ$ . This does not violate the way that arcsine was defined — arcsine was defined using the restricted domain $[-\pi/2,\pi/2]$ , or $[-90^\circ, 90^\circ]$ in degrees.

Second, let’s look at drawing angles in standard position. The angle in the second quadrant is clearly the reflection of the angle in the first quadrant through the $y-$ axis.

Third, let’s use a trigonometric identity to calculate $\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right)$ :

$\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right) = \sin \pi \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) - \cos \pi \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$=0 \cdot \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) + 1 \cdot \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)$

$= \displaystyle \frac{1}{3}$

Fourth, and perhaps most convincingly for modern students (to my great frustration), let’s use a calculator:

All this to say, blinding computing $\sin^{-1} \frac{1}{3}$ uses incorrect logic when solving this problem.

Tomorrow, we’ll examine what happens when we try to solve the triangle using these two different solutions for $\gamma$ .

Inverse functions: Arcsine and SSA (Part 13)

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.

Inverse functions: Arcsine and SSA (Part 12)

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.