Trig Ratios as Percents

My audience is mostly folks interested in math education in one way or another, so it’s no use starting this post off with “All you may know about trigonometry ratios is likely captured in the gibberish mnemonic SOHCAHTOA.” Your understanding of trigonometry ratios is no doubt more sophisticated than that.

But have you thought about trig ratios as percents? This will be enough for most of you:

sin θ = \(\mathtt{\frac{opposite}{hypotenuse} = \frac{?}{100}}\) = percent of hypotenuse length

It makes sense when you dredge up the 6th-grade math you remember and start making connections between it and the trigonometric ratios sine, cosine, and tangent (for example). After all, opposite : hypotenuse is the sine ratio, but it’s also just a ratio. If we think of it as a percent, we could say that if the sine of a reference angle is equal to 0.75, that means that the side opposite the angle in a right triangle is 75% the length of the hypotenuse. If the cosine were 0.75, that would mean that the side adjacent to the reference angle is 75% the length of the hypotenuse, since cosine is the ratio adjacent : hypotenuse. And a tangent of 0.75 means that the opposite side is 75% the length of the adjacent side, because tangent is simply the ratio opposite : adjacent.

The percent connection (or fraction; doesn’t have to be percent) strikes me as being immediately more useful for seeing meaning in values for trigonometric ratios. They usually go by students as just values which can’t be put into a sentence—a long list of changing decimals in a lookup table. Yet, the percent connection is right there, waiting for us to combine our middle school math knowledge with new material. We could model what this process of meaning-making actually looks like, rather than just ask them to go make meaning and hope for the best.

Of course, it also helps to be able to visualize what a sine of 0.75 looks like. Try, say, \(\mathtt{49^\circ}\) below on the unit circle and press Enter. That gives me something that looks pretty close to a sine of 0.75 (an opposite side that is \(\mathtt{\frac{3}{4}}\) the length of the hypotenuse, right?).

  θ = °

cos-sin-1
1-tan-sec
cot-1-csc

But the interactive tool, while helpful maybe, isn’t necessary, I don’t think. One can think about drawing a right triangle, say, with an adjacent side length about 80% of the hypotenuse length (a cosine of about 0.8). It will have to be longer than it is tall, relative to the reference angle, to make that work. The percent connection thus links a trigonometry ratio value to a simple and accessible visual.

An Example Problem: Testing Out the Percent Connection

41° 96 x

The basic mathematical (as opposed to contextual) trigonometry practice problem looks like this: Determine the length of \(\mathtt{x}\).

I can’t say the percent connection makes this a faster or more efficient process. What I would say is that knowing that the sine of 41° means the percent of the hypotenuse length represented by the opposite side length makes me feel like I know what I’m doing, other than moving numbers and symbols around. (Thinking about percents also gives us a way to estimate what my \(\mathtt{x}\) will be, if I know that the figure is drawn to scale.)

The sine of 41° is approximately 0.65605902899. With the percent connection, I know that this means that the opposite length is about 65.61% the length of the hypotenuse. It’s hard to overstate, I think, how useful it is to be able to wrap all of this number-and-variable work into one sentence like this: 96 is about 65.61% of x. I can climb the last few steps from there, by either dividing or setting up an equation—however the work happens, I at least have some background meaning to the numbers I’m playing with.

We can continue from there, of course (as we can without the percent connection, but so rarely do because the tedium of setting up and solving for the variable has overloaded us). The tangent of 41°, approximately 0.86928673781, tells us that the opposite side is about 86.93% the length of the adjacent side.

This guy gets it, and he seems to be the only one. It shouldn’t come as any surprise that he’s an experienced mathematics teacher a computer scientist who’s never taught. But, you know, it really should surprise us. Someday.


trigonometry

Sequencing Props

Last time (here), we talked about III.27 from The Elements, which told us that “in equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences.” Here we’ll go over the proof of the previous proposition in the sequence, III.26, which is different from III.27 in ways not easy to access based on just the description: “in equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences.” So . . . let’s take a look.

sequence

We start by drawing some things and making some assumptions. First, Circles \(\small\mathtt{ABC}\) and \(\small\mathtt{DEF}\) are congruent, the central angles (\(\small\mathtt{\color{green}{\measuredangle{BGC}}}\) and \(\small\mathtt{\color{green}{\measuredangle{EHF}}}\)) are congruent, and the inscribed angles (\(\small\mathtt{\color{blue}{\measuredangle{BAC}}}\) and \(\small\mathtt{\color{blue}{\measuredangle{EDF}}}\)) are also congruent. What we want to show is that the minor arc \(\small\mathtt{BC}\) is congruent to minor arc \(\small\mathtt{EF}\).

Chain of Begats

Euclid proceeds by drawing \(\overline{\small\mathtt{BC}}\) and \(\overline{\small\mathtt{EF}}\), which I’ve already done above. He then concludes that \(\overline{\small\mathtt{BC}}\) and \(\overline{\small\mathtt{EF}}\) are congruent by reasoning as follows (using SAS, or I.4, which we covered here):

  1. Circles \(\small\mathtt{ABC}\) and \(\small\mathtt{DEF}\) are congruent, so their radii are congruent.
  2. This means that \(\overline{\small\mathtt{BG}}\) is congruent to \(\overline{\small\mathtt{FH}}\) and \(\overline{\small\mathtt{CG}}\) is congruent to \(\overline{\small\mathtt{EH}}\).
  3. The central angles (\(\small\mathtt{\color{green}{\measuredangle{BGC}}}\) and \(\small\mathtt{\color{green}{\measuredangle{EHF}}}\)) are congruent by assumption.
  4. Thus, \(\overline{\small\mathtt{BC}} \cong \overline{\small\mathtt{EF}}\) via SAS.

The Peril of Working Backwards

sequence

The sequence of the propositions in the Elements is important, which we’ll feel right now when I tell you that the next step in this proof relies on a previous theorem, III.24. That proof, which we’ll have to get to later, shows that when two circles have arcs (“segments,” actually) that lie on congruent line segments, those two arcs are also congruent. (There’s also some weirdness in the definition of “similar segments” used in that proof [and this one] which we may want to think about later too.)

For our purposes, this means that major arc \(\small\mathtt{BC}\) is congruent to major arc \(\small\mathtt{EF}\). And since the two entire circles are congruent by assumption, this means that the minor arcs must be congruent too.

So, in III.27, we showed that the angles are equal when we are provided a given that the intercepted arcs are equal. In this proof, we show that the arcs are equal when we are provided a given that the angles are also.

Sequence Is Important to Think About, Not Necessarily to Follow

sequence

Order is something I definitely think a lot about when it comes to designing curricula. And our public conversation about sequencing in mathematics education strikes me as incredibly naive. What I have to listen to more often than not is that students “need X to do Y.” I mentioned above that a proof of III.24 was used to construct a proof of III.26, but it does not necessarily follow that III.24 is required to create this proof (it might be required, though). Needing X to do Y is often defended on similarly shallow evidence. Usually, the foundation for this argument is simply the sequence that the arguer is used to. Or a shallow understanding of a topic.

While it is certainly true that some topics must appear before others, we should also think about what benefits to learning playing around with order can bring. At the moment, what seems to drive almost all of our sequencing decision-making is tradition, ignorance, and fear.


Image credit: pratanti

Ceva’s Theorem

The structure of this proof, a proof of Ceva’s Theorem, comes from a neat little book called Geometry Revisited. But I’ll start with something the authors of that book ask us to recall: “that the areas of triangles with equal altitudes are proportional to the bases of the triangles.”

Canvas not supported.

The equation below the left triangle shown is of the form \(\small\mathtt{A_1 = b_1 \cdot k}\) and the equation for the triangle to the right is \(\small\mathtt{A_2 = b_2 \cdot k}\), where \(\small\mathtt{A}\) stands for area (in square pixels) and \(\small\mathtt{b}\) stands for base length (in pixels). Because the two triangles have equal heights (equal altitudes; in this case 180 pixels), we can use the constant \(\small\mathtt{k}\) to represent \(\small\mathtt{\frac{1}{2} \cdot h}\) in each area equation.

You can drag the bottom left vertex of the leftmost triangle or the bottom right vertex of the rightmost triangle to change the base length of either triangle. Note that the ratio of the areas is equal to the ratio of the base lengths (i.e., “the areas of triangles with equal altitudes are proportional to the bases of the triangles”). Since we have \(\small\mathtt{A_1 = b_1 \cdot k}\) and \(\small\mathtt{A_2 = b_2 \cdot k}\), then \(\small\mathtt{\frac{A_1}{b_1}}\) gives us the same number as \(\small\mathtt{\frac{A_2}{b_2}}\), namely \(\small\mathtt{k}\).

What the Theorem Says and a Proof

ceva's theorem

Draw three line segments \(\overline{\small\mathtt{AX}}\), \(\overline{\small\mathtt{BY}}\), and \(\overline{\small\mathtt{CZ}}\) inside a triangle, each from a vertex to the opposite side. Ceva’s Theorem (Ceva is pronounced CHEH-vuh) says that if the line segments meet at a point (are concurrent), then \(\small\mathtt{\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB}}\) = 1.

So, let’s see. Given what we learned above, \(\small\mathtt{\frac{BX}{XC}}\) = \(\mathtt{\frac{A_{ABX}}{A_{AXC}}}\). Each of these ratios is also equal to \(\mathtt{\frac{A_{PBX}}{A_{PXC}}}\).

That last ratio is also equal to \(\mathtt{\frac{A_{ABX} – A_{PBX}}{A_{AXC} – A_{PXC}}}\), because we can safely subtract across the numerators and then the denominators when working with equal part-to-part ratios (can you work out why?). Finally, subtracting the area of \(\small\Delta\)\(\small\mathtt{PBX}\) from the area of \(\small\Delta\)\(\small\mathtt{ABX}\) gives us the area of \(\small\Delta\)\(\small\mathtt{ABP}\), or \(\small\mathtt{A_{ABP}}\). Similarly, subtracting the area of \(\small\Delta\)\(\small\mathtt{PXC}\) from the area of \(\small\Delta\)\(\small\mathtt{AXC}\) gives us the area of \(\small\Delta\)\(\small\mathtt{CAP}\), or \(\small\mathtt{A_{CAP}}\). We have wound up with a chain of equals signs to prove Ceva’s Theorem, giving us, in the end, \(\small\mathtt{\frac{BX}{XC}}\) = \(\small\mathtt{\frac{A_{ABP}}{A_{CAP}}}\).

So now we do that with the other two sides. Here are all three equals-sign chains:

ceva's theorem

  1. \(\small\mathtt{\frac{BX}{XC}}\) = \(\mathtt{\frac{A_{ABX}}{A_{AXC}}}\) = \(\mathtt{\frac{A_{PBX}}{A_{PXC}}}\) = \(\mathtt{\frac{A_{ABX} – A_{PBX}}{A_{AXC} – A_{PXC}}}\) = \(\mathtt{\frac{A_{ABP}}{A_{CAP}}}\)
  2. \(\small\mathtt{\frac{CY}{YA}}\) = \(\mathtt{\frac{A_{BCY}}{A_{BYA}}}\) = \(\mathtt{\frac{A_{PCY}}{A_{PYA}}}\) = \(\mathtt{\frac{A_{BCY} – A_{PCY}}{A_{BYA} – A_{PYA}}}\) = \(\mathtt{\frac{A_{BCP}}{A_{ABP}}}\)
  3. \(\small\mathtt{\frac{AZ}{ZB}}\) = \(\mathtt{\frac{A_{CAZ}}{A_{CZB}}}\) = \(\mathtt{\frac{A_{PAZ}}{A_{PZB}}}\) = \(\mathtt{\frac{A_{CAZ} – A_{PAZ}}{A_{CZB} – A_{PZB}}}\) = \(\mathtt{\frac{A_{CAP}}{A_{BCP}}}\)

Then just cross out stuff to see that \(\mathtt{\frac{A_{ABP}}{A_{CAP}} \cdot \frac{A_{BCP}}{A_{ABP}} \cdot \frac{A_{CAP}}{A_{BCP}}}\) = 1, which means that \(\small\mathtt{\frac{BX}{XC} \cdot \frac{CY}{YA} \cdot \frac{AZ}{ZB}}\) = 1.

Huzzah!