SP #7: Unit Q: Concept 2

“Please see my SP7, made in collaboration with Marisol R. by visiting their blog here.  Also be sure to check out the other awesome posts on their blog”

I/D #3: Unit Q Concept 1- Pythagorean Identities

1. INQUIRY ACTIVITY SUMMARY: 
   

   
   

   1.     An identity is a proven fact and formula that are always true. The Pythagorean Theorem is an identity because it has been proven that two sides of a right triangle squared is equal to the hypotenuse squared(a^ + b^2= c^2). Using the variables x,y, and r, the the pythagorean theorem is x^2+ y^2 = r^2. Seeing that (x/r)^2 + (y/r)^2=1, and the ratio of x/r is cosine and y/r is sine signifies that sin2x+cos2x=1 relates to both trig functions (sin and cos). A right triangle on the first quadrant of the Unit Circle also has the coordinates sin and cos so using the Pythaorean theorem you can use that to obtain sin2x+cos2x=1. Using the ordered pair for 30* is (rad3/2,1/2), so (rad3/2)^2 + (1/2)^2 = 1 shows that this identity is true.

   
   

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1. INQUIRY ACTIVITY REFLECTION
   
   
   
   1. “The connections that I see between Units N, O, P, and Q so far are…” The Unit Circle and its ratios play a vital part in the Pythagorean Identities. the Pythagorean Theorem plays together with the Pythagorean Identites and the ratios from the Unit Circle together with their trig ratios.
   
   
   
   
   
   2. “If I had to describe trigonometry in THREE words, they would be…”ratios, Unit Circle, Pythagorean Theorem (these can be 3 separate words or 3 words that go together… ideally I would like 3 separate words though)

WPP #13 & 14.

This WPP13-14 was made in collaboration with Marisol Reyes.  Please visit the other awesome posts on their blog by going here.

The Problem:

Law of Sines 

1.Susie and Barbara plan to meet at Boudin Bakery SF for lunch. While Susie walks toward the cafe, she sees that Barbara is across from her in the parking lot at a distance of 22 feet. Barbara is N27*E from Boudin while Susie is N62*W from the bakery. What is the distance between Barbara and Boudin Bakery?

Law of Cosines 

2.) The two meet up at Boudin Bakery then decide they want to go to the beach. Both of them leave from the same point. Susie is at a bearing of 034° and drives at 30 mph while Barbara is at a bearing of 238° and drives at 52 mph. If they drive for 2 hours, what is the distance between them? 

 (http://data1.whicdn.com/images/27749337/large.jpg)

(http://static3.refinery29.com/bin/entry/038/x/153691/boudin-bakery-chowder-bowl-3x4.jpg)

 

The Solution

 Solution to #1:

 Solution to #2:

 

BQ#1 - Unit P

2. Law of Sines- Why is SSA ambiguous? SSA is ambiguous because we are only given one angle so we don't know how the triangle looks like since there could be two possible triangles, one, or even no possible triangle because we are only given one angle. In the example below we are only given angle A therefore we have to start the problem assuming there could be two possible triangles.
                    
 (an example of one possible triangle example)

4. Area Formulas- How is the “area of an oblique” triangle derived?  The area of an oblique triangle is derived by using the formula of one half of the product of the two adjacent sides given and the sine of their included angle. (Ex. Area=1/2absinC).  In the example below, if you plug in the values you have:
Area=1/2(16)(12)*sin(102). Make sure your calculator is in degrees mode and you get 95.5 u^2.

                                              Example of an oblique triangle:

(http://image.mathcaptain.com/cms/images/41/oblique-triangle-solved-problems.jpg)
  
How does it relate to the area formula that you are familiar with?

The area formula I am familiar with is Area= (length)(width)(height). The formula for the area of an oblique triangle relates to it because we are measuring A=1/2absinC where the height of the triangle is multiplied by (1/2) times the width (b) times the length (a).  

References:
 http://image.mathcaptain.com/cms/images/41/oblique-triangle-solved-problems.jpg

WPP #12: Unit O: Concept 10 : Solving angle of elevation and depression word problems

a. Barbara is about to ski down a mountain in Oregon. She estimates the angle of depression from where she is now to the finish line of the course to be 22*. She knows that she is 400 feet higher than the base of the course. How long is the path that she will ski? (Round to the nearest foot).


b. Barbara forgot her skis at the top of the mountain so she needs to trek back up to retrieve them! From where she stands at the finish line, the angle of elevation to the top of the mountain is 22*4'. If the base of the mountain is 454 feet from Barbara, how high is the mountain (to the nearest foot) ?

 (http://freeskier.com/wp-content/uploads/2013/10/Red_1-1024x638.jpg)

Solution:

 Angle of Depression:

Angle of Elevation:
 

I/D2: Unit O Concepts 7-8 - How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY

1.Cutting the 30-60-90 triangle in half gives us the 30 degree angle measurement since 60 divided by 2 is 30. To get the height note that we know what the hypotenuse is 1 since each side is 1 knowing it is an equilateral triangle and the base is 1/2 since half of one side is 1/2 and we need to find height so use the Pythagorean Theorem and you end up with rad3/2. To translate each value into normal values (without fractions) simply multiply each value by 2 so you end up with the height as rad3, the base as 1 and the hypotenuse as 2. Using n means we know each side shares the same pattern for the triangle. 

2. The 45-45-90 triangle is taken from a square and cut diagonally to get the 45 degree measurements from the triangle. 90 degrees cut in half is 45. To get the  hypotenuse you use the Pythagorean theorem and plug in 1 for the adjacent and opposite side since we know each side is labeled with 1 so we solve for c. we end up with rad2 for the hypotenuse. Using "n" means we use this variable to note that the relationship of each side is equal. 

INQUIRY ACTIVITY REFLECTION

“Something I never noticed before about special right triangles is…” Something I never noticed before about special right triangles is they each have rules that contain patterns each time you solve one!

“Being able to derive these patterns myself aids in my learning because…” Being able to derive these patterns myself aids in my learning because it gives me a greater understanding of the rules for special right triangles and understand where the constant n comes from and how it plays a part in each triangle. 

I/D #1: Unit N Concept 7: How do SRTs and UCs relate?

Inquiry Activity Summary:

1. The 30*  triangle has sides labeled (r,x,y). "r" stands for the hypotenuse; "x" adjacent or horizontal value; and "y" opposite or vertical value. After labeling the triangle according to the rules of Special Right Triangles, (r equals 1 from 2x/2x. y equals 1/2 from x/2x. x equals rad3/2 from rad3*x/2x.) Take note that the three sides are simplified so that the hypotenuse equals 1 (the radius of the Unit Circle is always equal to 1). Then, the hypotenuse was labeled with "r", the adjacent side with "x" and the opposite with "y". Next, I drew a coordinate plane, origin being located at the labeled 30 degrees (0,0) (so the triangle could lie in quadrant 1). and all three vertices labeled as ordered pairs: 60*  (1/2, rad3/2) and the last vertice 90* as (rad3/2,0) .


 













2. The 45* triangle was labeled according to the rules of SRT, taking note that each value was divided by rad2 so that the hypotenuse equals 1 (the radius of the Unit Circle is always equal to 1) (longest side, r, stood for rad2/rad2 equal to 1, 1/rad2 equal to rad2/2 and 1/rad2 equal to rad2/2). I labeled the hypotenuse "r", horizontal value "x", and vertical value "y". Next, I drew the coordinate plane labeling x and y axis and the origin of (0,0)  at the labeled 45*, the next point at (rad2/2, rad2/2) and the last at (rad2/2, 0). 




 3. The 60* triangle was labeled according to the rules of SRT dividing each side by 2x so that the hypotenuse equals 1 (the radius of the Unit Circle is always equal to 1), ("r" equal to 1 from 2x/2x; "y" equal to rad3/2 from xrad3/2x; "x" equal to 1/2 from x/2x). Next, I labeled the hypotenuse "r", horizontal value "x" and the vertical value "y". Lastly, I drew the coordinate plane labeling the x and y axis, origin at (0,0) and the corresponding ordered pairs per vertice (at 30*: (1/2,rad3/2) and (1/2,0) at 90*. 

 












4. This activity helped me derive the Unit Circle because it allows you to visually take note of where the angles and ordered pairs from the UC come to be and how each are solved for.  For example, taking a closer look at the 45* triangle, we could note where its ordered pair came simplifying each side using the rules of SRT. 

 (http://jwilson.coe.uga.edu/EMAT6680Su12/Jackson/Writeup10DMJ/Unit%20Circle.PNG)

5.  The quadrant drawn in this activity lies in the first quadrant (note all four quadrants in Figure 2). If you draw the 30* triangle in quadrant 2 it's csc/sin are positive, sec/cos and tan/cot negative are negative. In the third quadrant, tan/cot are positive, and csc/sin, cos/sec are negative. Lastly, note that cos/sec are positive and csc/sin, tan/cot are negative.

 (http://www.google.com/imgres?client=firefox-beta&hs=hI5&sa=X&rls=org.mozilla%3Aen-US%3Aofficial&channel=sb&biw=1280&bih=673&tbm=isch&tbnid=1aKd0mo1Wb16-M%3A&imgrefurl=http%3A%2F%2Fwww.sparknotes.com%2Fmath%2Ftrigonometry%2Ftrigonometricfunctions%2Fsection3.rhtml&docid=3IGWql635sbzWM&imgurl=http%3A%2F%2Fimg.sparknotes.com%2Ffigures%2F0%2F067486b8a9659518b7099dac07405d29%2Fquadrantsigns.gif&w=210&h=210&ei=UlUJU5DHF7LlygGK8YGQDw&zoom=1&ved=0CFcQhBwwAQ&iact=rc&dur=442&page=1&start=0&ndsp=17) 

 

(https://encrypted-tbn0.gstatic.c/images?q=tbn:ANd9GcTokzujb26VtG3kOJjr5Hiq9rR1O7TWjkroQ8gv_peFYJc6oBNBUA)

Inquiry Reflection Activity: 

1. The coolest thing I learned from this activity was learning a basis for trigonometry using the rules for SRTs and UCs! Seeing how the curriculum for geometry and trigonometry tie together is fascinating!
2. This activity will help me in this unit because finding the value for the trig functions are more easily solved if you understand the unit circle in more depth. 
3. Something I never realized before about special right triangles and the unit circle is how they related with each other and the components that are used carefully to calculate each angle and degree that make up the unit circle.

References:

• http://jwilson.coe.uga.edu/EMAT6680Su12/Jackson/Writeup10DMJ/Unit%20Circle.PNG
• https://encrypted-tbn0.gstatic./images?q=tbn:ANd9GcTokzujb26VtG3kOJjr5Hiq9rR1O7TWjkroQ8gv_peFYJc6oBNBUA
• http://www.google.com/imgres?client=firefox-beta&hs=hI5&sa=X&rls=org.mozilla%3Aen-US%3Aofficial&channel=sb&biw=1280&bih=673&tbm=isch&tbnid=1aKd0mo1Wb16-M%3A&imgrefurl=http%3A%2F%2Fwww.sparknotes.com%2Fmath%2Ftrigonometry%2Ftrigonometricfunctions%2Fsection3.rhtml&docid=3IGWql635sbzWM&imgurl=http%3A%2F%2Fimg.sparknotes.com%2Ffigures%2F0%2F067486b8a9659518b7099dac07405d29%2Fquadrantsigns.gif&w=210&h=210&ei=UlUJU5DHF7LlygGK8YGQDw&zoom=1&ved=0CFcQhBwwAQ&iact=rc&dur=442&page=1&start=0&ndsp=17 
• http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/8c85deff-b9ed-4c80-8b44-16e8e9322af2.png
• https://encrypted-tbn0.gstatic.c/images?q=tbn:ANd9GcTokzujb26VtG3kOJjr5Hiq9rR1O7TWjkroQ8gv_peFYJc6oBNBUA