I/D #3: Unit Q Concept 1: Pythagorean Identites

INQUIRY ACTIVITY SUMMARY:
1. cos2x+sin2x=1 is derived from the unit circle. The pythagorean theorem is x^2+y^2=r^2. In the picture below I demonstrate how the pythagorean theorem relates to cos2x+sin2x=1, since the ratio for sine is y/r and for cosine its x/r we can also substitute that into the Pythagorean theorem and then divide everything by r2 to make it equal to one. This is another method for figuring out how cosine and sine are derived from the Pythagorean theorem.

2. We can also derive other forms of the equation sin2x+cos2x=1:

What I did in the first example is divided everything on both sides by sin2x. We then discovery that the division of sin2x leads to some ratio and reciprocal identities that change the equation to cotx and cscx. For step 3 I just substituted the identities into the equation. 

For the second example I started off by dividing by cos2x. This also resulted in a ratio and reciprocal identities. Last step I substituted the new identities into the equation.

INQUIRY ACTIVITY REFLECTION:
1.The connections I see between Unit N, O, P, and Q so far are that many different equations can be derived from the unit circle and they can be  right triangles or non right triangles.
2.If I had to describe trigonometry in three words, they would be Unit Circle, Triangles, and SOHCAHTOA.

WPP #12&13: Law of Sines and Cosines Applications

This WPP 13-14 was made in collaboration with Cynthia A.  Please visit the other awesome posts on their blog by going here
a)Michael and Harry are 4 miles apart when they see a stranded cat on a tree. Michael looks up N35E in the direction of the cat, Harry looks N51W at the cat. What is the height of the cat on the tree from the midpoint?

Solution:

b)Michael and Harry then head their own ways from the tree. By the time try leave it's 2pm. Michael is headed at 027 degrees and walking at the pace of 2.2 miles per hour. Harry is moving at the pace of 2.7 miles per hour at the bearing of 118 degrees. How far apart will they be by 4pm?

BQ #1: Unit P Concept 2 Law of Sines & Concept 4 Area Formulas

2.Law of Sines
SSA is ambiguous because we can have one solution, two solutions, or no solution depending on the existing rules of Sines and triangles.

In this first case we have two solutions. This is possible in this type of problem because with the information given we were able to use the Sine of A to solve for the missing pieces. The first thing i did was solve for SinC which equaled to .1573 then took the inverse which gave us the value of the angle of C equal to 9. This problem did not meet any wall that would go against the laws of triangles or sines. It met the requirements of staying within 180 degrees and it Sine was not larger than 1. 

This case deals with one solution. The first thing I did was use Sine B as the main factor to solve for the missing pieces which in this case I solved for the Sine of A. After taking the inverse of SinA I got 8.1 as the angle of the first triangle, then i subtracted 8.1 from 180 to get the value of angle A for the second triangle and this is where I met a wall. Angle B has a value of 101 degrees, there can only be one obtuse angle within a triangle. When I added 101 to 171.9 it went over 180 degrees of a triangle so this meant there was only one solution.

Last example is no solution. This we automatically knew was not going to have any solutions because once we took the SinA it was bigger than one and sine cannot be bigger than 1. 

4.Area Formulas

The area of oblique triangles originated from the area formula of a triangle which is A=1/2bh. We derived it by for example in the triangle below:

http://www.compuhigh.com/demo/lesson07_files/oblique.gif

We took the sinC of triangle number 1 and made it equal to h/a also known as (O)pposite/(H)ypotenuse since were dealing with sine. Then we cross multiplied and got that h=asinC, this is the first part of the total equation now we know the value of h. Next we had to substitute the value of h into the original equation of area of a triangle. It related to the Area formula that I know because it doesnt matter what angle you use, the values will always be consistent. It just depends on the angle in which you are trying to find the area for.

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhMuKQl55_kU_UW-3B15iIKWQnm4jTq9kMH-dZjUZuvtX0xxXTAW8NgvHuSp-NNhckfbfuyplyEFb-r6krVc1zrdAOZMcVPI4lYEQN25jdRx-RL48qsF4Q0-o46Bs20KEofChd0GjczZk/s400/hi.bmp

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

a)Michael measures from ground level the angle of elevation to the top floor of the concert stadium to be 14 degrees. If the base of the floor is 500 feet from the seats, how high are the top row seats?

b)Michael now stands up on the top floor of the concert stadium. He measures the angle of depression from the top floor to the base of the stage (10 feet above ground level) to be 16 degrees. He knows that he is now approximately 115 feet on the top of the top floor. How far are you horizontally from the stage?

http://31.media.tumblr.com/58dbe2cbb5f11655cbfb64d49415cc10/tumblr_mtb3d2e8lA1sqc3bro7_500.jpg

Solution:

I/D #2: Unit O Concept 7-8: Using the 30-60-90 and 45-45-90 Triangles

Inquiry Summary Activity:

To derive the Special Right Triangles from an equilateral triangle and a square, we had to arrange the values of each side to fit accordingly to the rules of the 30-60-90 and 45-45-90 triangles. The Pythagorean Theorem was used to change the value of the sides, when all sides were initially equal to 1.

1) 30-60-90 Triangle

First, I started off by labeling each side of the equilateral triangle 1. Next I cut the triangle down the middle, creating two 30-60-90 triangles. Next I labeled the 90 degree angle for each triangle. The 30 degree triangle is the top angle and the 60 degree angle are on the bottom ends. Next, I had to solve for the opposite side of the 60 degree angle. What we knew so far was that the hypotenuse (across from 90 degree angle) had a value of 1. The TOTAL value of the base of the triangle was equal to 1, after we cut it down the middle, it made each side of the bottom of the triangle equal to 1/2; so we knew the value of the opposite side of the 30 degree triangle was equal to 1/2. Now it was time to solve for the hypotenuse, so we use the Pythagorean Theorem. I set the missing side as a, then solved.

http://www.themathpage.com/aTrig/30-60-90-triangle.htm

In the Special 30-60-90 triangle rules, the value of the hypotenuse has to be equal to 2. So, this means since we got radical 3/2 for the side we solved for, we multiply all the sides of the triangles by 2. This will change out values to be: opposite to 90 degrees is 2, opposite to 60 degrees is radical 3, and opposite side to 30 degrees is equal to 1. Lastly, we add an n to the values of the 30-60-90 sides to help us solve for the missing side on the shared triangle. The variable allows us to solve for the "missing".


2)45-45-90 Triangle

After I labeled all sides of the square equal to 1, I cut the square diagonally through the middle. This created (2) 45 degree triangles. I labeled the 90 degree angle for each triangle to clearly see the 45 degrees on each triangle. The side opposite to the 90 degree angle is the hypotenuse. We don't know the value of the hypotenuse, we only have the value of the adjacent and opposite sides of the 45 degree angles; in order to figure out the hypotenuse value we have to use the Pythagorean Theorem.

 

 

 

http://www.getgeometry.com/images/45-45-90-triangle-ratios.png

 

 

Once we have used the Pythagorean Theorem, we get the hypotenuse to equal radical 2 for a 45-45-90 triangle. We then add the variable "n", so we have something to solve for. The purpose of the variable is so that we can solve for a missing side on the other triangle when the two are sharing a side; when we know all the sides to one triangle, it gives us the value of the shared side between the two so we can solve for the missing side on the second triangle.


Inquiry Activity Reflection:
1.Something I never noticed about special right triangles before is: that you can form a 30-60-90 or a 45-45-90 from shapes that don't seem to have triangles of this kind at first sight.
2.Being able to derive these patterns myself aids in my learning because: I can solve for any shape as long as I can derive special right triangles from them.