Dabbing and the Pythagorean Theorem

I enjoyed this article from Fox Sports. Apparently, a French Precalculus textbook created a homework problem asking if football (soccer) superstar Paul Pogba is doing the perfect dab by creating two right triangles. 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.

Proving theorems and special cases (Part 11): The Law of Cosines

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

2. Theorem. In $\triangle ABC$ where $a = BC$, $b = AC$, and $c = AB$, we have $c^2 = a^2 + b^2 - 2 a b \cos (m \angle C)$.

This is typically proven using the Pythagorean theorem:

Lemma. In right triangle $\triangle ABC$, where $\angle C$ is a right angle, we have $c^2 = a^2 + b^2$.

Though it usually isn’t thought of this way, the Pythagorean theorem is a special case of the Law of Cosines since $\cos 90^\circ = 0$.

There are well over 100 different proofs of the Pythagorean theorem that do not presuppose the Law of Cosines. The standard proof of the Law of Cosines then uses the Pythagorean theorem. In other words, a special case of the Law of Cosines is used to prove the Law of Cosines.

Another proof of the Law of Cosines 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).

Engaging students: Law of Cosines

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Allison Metlzler. Her topic, from Precalculus: the Law of Cosines. What interesting (i.e., uncontrived) word problems using this topic can your students do now?

Real world word problems are an effective engagement because the students can actually relate to the events occurring in the problem. Below are two word problems where one deals with animal footprints and the other talks about trapeze artists.
1. Scientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180 degrees, the more efficiently the organism walked. Based on the diagram of dinosaur footprints, find the step angle B.
2. The diagram shows the paths of two trapeze artists who are both 5 feet tall when hanging by their knees. The “flyer” on the left bar is preparing to make hand-to-hand contact with the “catcher” on the right bar. At what angle (theta) will the two meet?
The problems were obtained from http://www.muhsd.k12.ca.us/cms/lib5/CA01001051/Centricity/Domain/547/Trig/13-6%20Law%20of%20Cosines.pdf. How could you as a teacher create an activity or project that involves your topic?

Activities are a great way to engage students. They require the students to explore the topic and make new discoveries. It can also benefit students who learn best by doing hands-on work. The activity, http://hilbertshotel.wordpress.com/2013/01/10/law-of-sinescosines-mapquest/ involves the law of sines, the law of cosines, and MapQuest. You will need a map of your school or just one of your school’s buildings. The students will then create triangles to figure out the length of different parts of the school. In order to do this, the students will have to use the law of cosines and sines. They will be able to measure the angles of the triangles using protractors. Then they can calculate the lengths of the sides of the triangles. You can then relate this activity to the real world job of surveyors. You would also need to point out to the students that because they are rounding their calculations of the distances and angles, there is a loss of accuracy. Also, you should note that in real life, surveyors would compute the distances using a different method in order to be completely accurate. This activity is very interesting and helps the students get a good understanding of the law of cosines. How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic?

A video is a great way to engage students because it’s visual and auditory which helps student understand concepts better. The video below uses Vanilla Ice’s song, Ice, Ice Baby, to introduce the law of cosines. I would play it from the start until1:51. At 1:51, the video starts introducing the idea of the law of sine. Besides just introducing the general idea of the law of cosines, it also shows how it’s derived from the Pythagorean Theorem. The video also clearly states that the Pythagorean Theorem only works with right triangles so that’s why we need the law of cosines- to help solve all triangles. It points out that you cannot only solve for a side of the triangle, but also the angles of the triangle. Another reason this video is engaging is that it is a well-known song that is catchy. Thus, the students will be able to remember the connection between the video and the concept of the law of cosines.

References:

Apply the Law of Cosines (n.d.). In MUHSD.k12. Retrieved April 4, 2014, from http://www.muhsd.k12.ca.us/cms/lib5/CA01001051/Centricity/Domain/547/Trig/13-6%20Law%20of%20Cosines.pdf

Dahl, M. (Producer). (2009). Law of Cosines Rap- Vanilla Cosines [Online video]. YouTube. Retrieved April 4, 2014, from http://www.youtube.com/watch?v=-wsf88ELFkk

Newman, J. (2013, January 10). Law of Sines/Cosines “Mapquest”. In Word Press. Retrieved April 4, 2014, from http://hilbertshotel.wordpress.com/2013/01/10/law-of-sinescosines-mapquest/