Chapter 2 Project - Modeling Profit

# CH 5 How to Do Determining the Number of Triangles in the Ambuguous case (SSA): A Contruction Exploration

In each of the three different cases, I constructed three triangles with a 120 degree angle but different relationships between the lengths of the sides.

In Case 1, where line BC was longer than line AB, I discovered that one triangle could be formed while keeping angle A at 120 degrees.
In Case 2, where line BC was equal to line AB, it was impossible for line segment AC to form since line segment BC was not long enough to connect with the dashed line. Therefore, no triangles were formed.
In Case 3, where line BC was shorter than line AB, the same problem in Case 2 arose. Line segment BC was not long enough to connect with the dashed line. Again, no triangles were formed.

In conclusion, I found out that in order for an obtuse triangle to be created, the following rules apply*:

• If BC > AB, then one triangle is possible.
• If BC = AB, then no triangles are possible.
• If BC < AB, then no triangles are possible.
*Note: The lettering scheme I use may be different than the lettering schemes used byothers. However, the same concepts I mention here still apply.
In my constructions:
• Line segment BC refers to the side opposite of the obtuse angle measuring 120 degrees
• Line segment AB refers to the leg of the angle measuring 120 degrees which touches line segment BC as mentioned above





These three cases here were based on acute triangles, and the relationships between the lengths of the sides and the height of the triangle. In each of the cases, in order for the triangle to still have an angle A measuring 30 degrees, point C of line segment BC would have had to connect with the line segment.

In Case 1, line segment BC was shorter than the triangle’s “minimum” height, so therefore no triangles were constructed. The dashed red circle I drew represents the fact that no matter where point C is placed, line BC will never connect with the dashed line.
In Case 2, line segment BC was equal to the height, so one triangle was constructed.
In Case 3, line segment BC was greater than line segment AB which was greater than the height. One triangle was possible.

In conclusion, in an acute triangle:
• If BC < h, then no triangles are possible.
• If BC = h, then one triangle is possible.
• If BC > AB > h, then one triangle is possible.



In Exploration 3, I drew a triangle where line AB was greater than line BC which was greater than the triangle’s height. I discovered that two triangles were possible while keeping an angle A measuring 60 degrees. The triangle in Exploration 3 is similar to the triangles in Exploration 2. If this triangle were an Exploration 2 triangle, then point C would be able to connect to the dashed line in two places, therefore creating two triangles that we can see here. Basically, there is enough room for line BC to exist in two places.
In conclusion, if in an acute triangle:
• If AB > BC > h, then two triangles are possible.

Questions
1. Explain why sin C is the same in both triangles in the ambiguous case (exploration 3). This is why the Law of Sines is also ambiguous in this case.
Let’s recall the values for the triangle in Exploration 3. Line AB equals 8 cm (in the Law of Sines formula, this would be side c), angle A equals 60 degrees, line BC equals 7 ½ cm (LoS formula: side a), and the height equals 7 cm.
The Law of Sines is ambiguous in this case since two different triangles will share part of the same Law of Sines formula since they have a congruent angle and two sides (otherwise known as the misleading SSA congruency conjecture). In the triangle in Exploration 3:
  sin 60 = sin C
     7.5           8
The Law of Sines here misleads us into thinking there is only one triangle, when two are possible. If one were to solve this equation they would only get one of the triangles. It would require one to solve the sin B/b part (which is where the ambiguity comes in), to get the full picture.
2. Explain from diagram 4 in Exploration 2 why a unique triangle is determined if BC < AB.
In acute triangles, a triangle can be constructed when BC < AB since point C of line BC has enough length to “connect” to AB. However, two triangles are not possible since unlike in Exploration 3, there is not enough “room” for point C to connect in two places without point C overshooting line AB.
3. What did you discover about the number of triangles you can make given SSA (two sides and an angle)? Summarize the "rules" for all cases.
Given SSA, the case in Exploration 3, two triangles can be made.
 
All in all:

• If in an obtuse triangle:




• If BC > AB, then one triangle is possible.


• If BC = AB, then no triangles are possible.


• If BC < AB, then no triangles are possible.




• If in an acute triangle:
  
  
  
  • If BC < h, then no triangles are possible.
  
  
  • If BC = h, then one triangle is possible.
  
  
  • If BC > AB > h, then one triangle is possible.
  
  
  • If AB > BC > h, then two triangles are possible.
  
  
  
  

Since other people use different lettering schemes, I will list how my line segments correspond to the Law of Sines formula:
• AB = c
• BC = a
• AC = b

Chapter 4 - Solving Problems with Trigonometry

CH 1 Project - Modeling with Functions and Regression Equations

For this problem, I will have to graph and predict the hourly earnings of Canadian production workers.

First, I will have to enter this data into the statistics tables on my calculator (STAT > Edit). I will enter the years into L₁ and the wages into L₂.




Then, I will find the linear regression equation. I can do this by going to STAT > CALC > LinReg(a+bx). I entered LinReg(a+bx) L₁,L₂,Y₁ so the equation in the list of equations when I hit the "y=" button. You can show Y₁ on your calculator by going to VARS > Y-VARS > Function > Y₁.



r is the correlation coefficient, or how closely this equation matches the data. Rounded to the nearest one thousandth, r is approximately 0.962. This value is appropriate for the model, although I would prefer a value closer to 1.

Next, I will find the quadratic regression, which is a model of the data using a quadratic equation. I can do this by using QuadReg, instead of LinReg(a+bx).




The value of R² (rounded to the nearest thousandth is .948) also suggests that this model is appropriate, but also that the linear model is better.

Lastly, I will graph these models. I turned on "Plot1" in the "y=" screen so the scatter plot will also be graphed.



The bold line is the quadratic model, and the squares are the original scatter plot. The viewing window I used here is: xmin=1971.3, xmax=2011.3, ymin=2.4254, ymax=33.7546.

The original problem here was to predict the wages of Canadian workers in 2010. Now that my models are set up, all I have to do is view the table (2ND > GRAPH).



The estimates here are slightly different from each other, with the linear model predicting $28.55 and the quadratic model predicting $30.55/hr.

The regression equations here fit the data relatively well as shown in the graph above. The linear equation was much better though, since it's correlation efficient, r≈.962, is closer to 1 than the quadratic equation's correlation efficient, R²≈.948.

Ch P - Graphing Functions

The parent function is x^3 (black). We can change this function by using translations. To translate vertically we must add to y. Here, I translated by 1 and got y=x^3 +1 (red). To translate horizontally I shifted 5 to the right and subtracted from x, resulting in (x-5)^3 (blue). Combining both will result in (x-5)^3+1 (green). As for reflections, I set x to -x. I got a reflection across the x-axis using y=-x^3 (purple). We can also combine reflections and translations, as seen in y=-(x-5)^3-1 (Sharpie). There will not be a reflection across the y-axis on the original function since that would have no effect.

Linear Equations - Chapter P Project

Chapter P Lesson 4 focuses on linear equations. Linear equations are straight lines modeled by the equation y=mx+b, where m and b are constants. m is the slope of the equation while b is the y-intercept. Linear equations intersect the y-axis at the y-intercept and the x-axis at the x-intercept. Unlike quadratic equations, linear equations only have one x-intercept. To create an equation from a slope and two points, we use the point-slope form of a linear equation which is: y-y₁=m(x-x₁).

In this project, I chose to find the appreciation of stock value. To do this, I will use linear equations to estimate how stock value increases over time.

About Vince

Background:

Education

Elementary: Bella Vista

Middle School: Edna Brewer

High School: Oakland High

Work and Volunteer

None

What I want to be:

Not sure, but something involving math, such as engineering.

Interests:

At home, I usually read, listen to music, or play video games. Sleeping in and staying up late are also other hobbies of mine.