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Thursday, March 20, 2014

BQ# 1: Unit P

2. Law of Sines
SSA is ambiguous because we don't know if we will have no triangles, one triangle, or two triangles. We are only given one angle in this case. Now thinking back to the Unit Circle, we should remember how the lines on the circle are given to us by using Law of Sines when the radius is not equal to 1 as it has been given to us in Unit N.



4. Area Formula
The base is b and h is the height of the triangle. Here are are not given the height, so we have to use arcsin of the angle which, in this case, is equal to h/a. We are using opposite over hypotenuse for whatever angle we are using; in this case, it is angle C. We then take sin of the angle times 1/2ab. However, sometimes it won't have a and b given, so we have to take what is given and make sure we haves its opposite angle. All the sides must be different, so just think of all the variables in the equation being different. Here are the other formulas we can use:
A= 1/2bcsinA
A=1/2acsinB
A=absinC


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


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

Wednesday, March 19, 2014

I/D3: Unit Q - Pythagorean Identities

Inquiry Activity Summary
The Pythagorean is here again. We first saw it in unit N in the unit circle, and it relates to trig identities as well. Using x, y, and r, we can show that. (x^2/r^2)+(y^2/r^2)=(r^2/r^2) which is also (x^2/r^2)+(y^2/r^2)=1. This can be simplified to (x/r)^2+(y/r)^2=1. sin2x+cos2x=1 comes from the unit circle. The ratio for cosine on the unit circle is x/r. The ratio for since is y/r. Here we are going to take one of the magic 3 ordered pairs: 30 degrees. cos30 is equal to radical 3 over 2 and sin30 is equal to 1/2. Radical 3 over 2 squared is equal to 3/4. 1/2 squared is equal to 1/4. 3/4+1/4=1 which is where you get the one on the right side. Since we squared these fractions, we have to show it in our identities also. This identity is actually the Pythagorean theorem moved around, hence it is also called the Pythagorean identity. Keep in mind that an identity is "a proven fact and formula that is always true." This tactic works with any degree from the magic 3.

We can also derive the secant and tangent from sin2x+cos2x=1. If you divide all of that by cos2x, you will end up with tan2x+1=sec2x. With tanx=sinx/cosx, multiply the left and right sides by themselves to get a squared value for both. tanx*tanx=(sinx/cosx)(sinx/cos). In the end you will get tan2x=sin2x/cos2x.

Inquiry Activity Reflection
  1. The connections that I see between Units N, O, P, and Q so far are that they all relate to triangle and they all use the six trig functions.
  2. If I had to describe trigonometry in THREE words, they would be challenging, connected, and unexceptional.

Tuesday, March 18, 2014

WPP#13-14: Unit P Concept 6 & 7 - Law of Sines/Cosines

This WPP 13-14 was made in collaboration with Tracey Pham. Please visit the other awesome posts on her blog by going HERE.

Thursday, March 6, 2014

WPP #12 Unit O Concept 10: Angle of Depression and Elevation

Elevation
Yoshi is filming his music video and wants to have his shot at the beach. For one of his shots, he is going to be standing at the top of a lighthouse while the camera will be on the floor 18 feet away from the lighthouse. The angle of elevation from the camera lense to the top of the lighthouse is 34*14'. What is the height of the lighthouse?

Answer: 12.25 m

Elevation

Depression
Yoshi is now on the other side of the lighthouse. He hears his fans cheering him on from below 36.8 m away and looks straight at them and waves. What is the angle of depression if he was standing on the same leveled lighthouse? (Hint: disregard the orange-ish lines labeling 12.25 m; the lighthouse from Yoshi to the floor is 12.25)

Answer: 34.24*
Depression


SHOW YOUR WORK



Tuesday, March 4, 2014

I/D2: Unit O - Derive the SRTs

INQUIRY ACTIVITY SUMMARY
As seen in the square, you have to cut it diagonally because that is how you can get two triangles. If they are equally bisected, they will form two 45-45-90 triangles. You get the hypotenuse by using the Pythagorean theorem (a^2+b^2=c^2). We are told that the side lengths of the square are 1. That means the side lengths of a and b of the triangle are 1 as well. You take (1)^2+(1)^2=c^2. That will give you c=radical 2; this is your hypotenuse. "n" means the variable used to find your sides. Since in a 45-45-90 degree triangle the sides corresponding to the 45 degree angles are the same, we can label them with "n." 





In the 30-60-90 triangle, I got it by bisecting an equilateral triangle down the center. Since the sides of the equilateral are equal to 1, I knew the hypotenuse was 1. The base is 1/2 since we cut the triangle in half. I then used Pythagorean theorem to find the height which gave me a=radical 3 over 2. This translates to the normal pattern because side a is n equal to n. The hypotenuse is double that. Here we see that side a (1/2) was multiplied by 2. (2)(1/2)=1; that is our hypotenuse. To find the height, we are most familiar with n radical 3. Our n in 1/2. We multiply that by radical 3. That gives us radical 3 over 2 as our height. We use n because that's the variable we are most familiar with when we first learned about triangles in geometry. In this problem the b is n, the hypotenuse is 2n and the height is n radical 3.



INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles is
that they can be made from squares and triangles. After learning the unit circle, I thought special triangles were only used in that.
Being able to derive these patterns myself aids in my learning because
I now better understand my last unit as well as understand where my values come from. Before I just had to memorize the values of each side without knowing how I got them.