Modeling in GeoGebra

Introduction

In this entry, I’m going to demonstrate the use of GeoGebra to estimate a value for a fairly tricky trigonometry problem, then illustrate how to find the value using trigonometry and an appeal to WolframAlpha. In so doing, I hope to also illustrate the eight basic standards for mathematical practice within Common Core.

Here is the problem, as it appeared in the G+ Mathematics community (6/20/23: I reconstructed the problem because the original graphic was lost to me; I believe this is correct):

Although it was posted with the caption “No need to use pencil”, this problem is more difficult than it first looks.

Since I didn’t have a precise strategy for solving this using trigonometry, I wanted to have a ballpark estimate to which I could compare an eventual answer. I find that GeoGebra is useful for this.

We are given the following conditions:

Point A is six units away from point B.
Point X is three units below point A.
Point X is two units away from point Y.
There are two right angles, and two right triangles, in the diagram.

My approach was to place point B at the origin, place point C on the x-axis (so that it can slide back and forth), and then place the other three points to satisfy all but the third condition. I could then slide point C around until that third condition was satisfied, at which point I’d have an estimate.

There are probably other ways to build the model, but this is a workable one.

Place a point at the origin and rename it to B; place another point on the x-axis and rename it to C. Then create point A with the following in the Input line:

A = (x[C], sqrt(36 - x[C]^2))

If you move point C back and forth, you can see the effect on point A.

Point X can be created with the following in the Input line:

X = (x[A], y[A] - 3)

Note that X and Y have to be capitalized; they’re not available as lower-case variable names.

Draw in some segments so we can see the progress. This gives us:

There is one remaining segment and point to place. First we want to make sure that ∠YXB is a right angle; we do this by inserting an “angle with given size”. When told to do this, GeoGebra creates a transformation of point B, called B’, that is rotated by 90° clockwise around X.

Next, create a segment between X and B’. Then create a point at the intersection of the two segments, and call it Y. Finally, hide B’ and XB’, create a new segment XY, and do some other clean-up. This gives us:

Adding a text box allows us to track XY easily, and gives us an estimate for BC when XY = 2:

Our target solution, then, is about 3.58.

Solving the Problem

There are a few approaches we could use to determine the length of BC given the initial conditions, but I could not find one so simple as to not require “pencil or paper”. Indeed, the approach I developed still involved me appealing to WolframAlpha.

Note that there are a total of five triangles in the diagram: △XYB, △XCB, and △CBA are right triangles, while △AYX and △AXB are not. Unfortunately, we know the most about the sides of the latter two, so we need to rely on the Law of Cosines to fill in the gaps.

For △AYX, we know that AX = 3 and YX = 2. Let y = AY. By the Law of Cosines, \[ a^2 + b^2 – c^2 = 2ab \cos{\gamma} \\ 3^2 + y^2 – 2^2 = 2(3)(y)\cos{A} \\ \Rightarrow y^2 + 5 = 6y \cos{A} \]

For △XYB, we know that XY = 2 and BY = 6 – y. By the Pythagorean Theorem: \[ XB^2 + 2^2 = (6 – y)^2 \\ \Rightarrow XB = \sqrt{(6 – y)^2 – 4} = \sqrt{36 – 12y + y^2 – 4} \\ = \sqrt{y^2 – 12y + 32} \]

For △AXB, we know that AB = 6 and AX = 3. We now have XB as well, so by the Law of Cosines: \[6^2 + 3^2 – \left(\sqrt{y^2 – 12y + 32}\right)^2 = 2(6)(3) \cos{A} \\ \Rightarrow 36 + 9 – y^2 + 12y – 32 = 36 \cos{A} \\ \Rightarrow 13 – y^2 + 12y = 36 \cos{A} \]

We have two values for cos(A): \[\cos{A} = \frac{y^2 + 5}{6y} \\ \cos{A} = \frac{13 – y^2 + 12y}{36}\]

Cross-multiplication gives us: \[(13 – y^2 + 12y)(6y) = (y^2 + 5)(36) \\ \Rightarrow (13 – y^2 + 12y)(y) = (y^2 + 5)(6) \\ \Rightarrow 13y – y^3 + 12y^2 = 6y^2 + 30 = 0 \\ \Rightarrow -y^3 + 6y^2 + 13y – 30 = 0\]

There is a formula for finding the roots of a cubic function, but it’s significantly more complicated than the Quadratic Formula, and often involves complex elements even if the actual value is real. This cubic has three real roots, each of which is indeed very messy, but which are approximately -2.7401, 1.5154, and 7.2246. Since 0 < y < 6, y ≈ 1.5154, and \(\angle A = \cos^{-1} \left(\frac{y^2 + 5}{6y}\right)\). Furthermore, \(\frac{y^2 + 5}{6y} \approx \frac{1.5154^2 + 5}{6 \cdot 1.5154} \approx 0.8025\), so \(\angle A \approx \cos^{-1} (0.8025)\).

Since we know AB and m∠A, we can determine BC: \[\sin{A} = \frac{BC}{AB} \Rightarrow BC = AB \sin{A} = 6 \sin{A} \\ \approx 6 \sin \left(\cos^{-1} (0.8025)\right) \approx 3.58\]

This is the same value that we found using GeoGebra.

Common Core Standards

In this entry, I have demonstrated several of the Common Core Standards for Mathematical Practice:

I made sense of the problem and found an entry point. In this case, my strategy was to use the shared vertex A and the Law of Cosines to calculate a value for m∠A, which I then used to find BC. I also started with a target value based on a model so that I’d have a sense of whether my trigonometric strategy was accurate.
I reasoned through different modalities: I used GeoGebra for a geometric representation, applied the Law of Cosines to develop a cubic equation, which I then approached algebraically before returning to trigonometry.
I constructed my argument by first showing how to approximate the solution in a trusted geometric modelling package, and then laid out the steps for a more rigorous solution.
While this problem has not been tied to a real-world situation, I used models and demonstrated both visual and algebraic approaches to the same problem.
I used GeoGebra and WolframAlpha strategically, applying the first to get an estimate and using the second to avoid the complicated mathematics behind the Cubic Formula.
I attended to precision, particularly by making sure to use the “approximately equal to” symbol as a reminder that the final answer is an approximation, and by striving to use an appropriate number of decimal places.
Throughout my explanation, I relied on the various properties (associative, distributive, and so on) to rearrange my equations into more usable forms.
I constructed my argument so that I could tell if my final answer was reasonable. I relied on shared sides and angle to construct my strategy.

Introduction