Wednesday, March 26, 2014

SP #7: Unit Q Concept 2: Finding all trig functions when given one trig function and quadrant

This SP #7 was made in collaboration with Anthony Lopez. Please visit the other awesome and non-cheesebuckety posts on their blog by going here.
**Also, both Anthony (cheesebucket) and I worked together on this post but someone *cough*Anthony*cough* has poor penmanship so in order to spare you the trouble of trying to decipher it, here it is in my writing. 


Identity Work


 What is this problem about? 
This quadrant discusses how to find ALL trig. functions when you are given one trig. functions and quadrant, using the identities from concept 1. It allows us to apply these fundamental identities to more difficult concepts such as this one.


What must the viewer pay special attention to in order to understand? 
You first have to decide the quadrant in which the values lie in by using ASTC quadrants, this will only determine the signs (very important!!) But also please remember you must using reciprocal, ratio, or Pythagorean identities to find the remaining values (don't rely on just SOHCAHTOA, we will use this to find our values.) AND you can only have ONE unknown trig function within solving the identity no matter what identities you use. Also make sure that you properly rationalize all your answers!!



SOHCAHTOA Work (Verification)


We use SOHCAHTOA to verify our answers from the identities we had previously used. We found the hypotenuse value by using the Pythagorean theorem and then simply used the values from the triangle in our ratios to verify that what we had originally gotten with the identities was correct. 




Wednesday, March 19, 2014

I/D #3: Unit Q: Pythagorean Identities

Inquiry Activity Summary

1. Where does sin ^2x + cos^2x = 1 come from?

Let's go back to oh like maybe three weeks?? (I'm awful with time) where we first learned about the Unit Circle. In the Unit Circle we learned that the ratio for cosine is (x/r) and the ratio of sine is (y/r). Now let's reflect back on Pythagorean Theorem which is a^2 + b^2 = c^2,  but in context of the Unit Circle it would be x^2 + y^2 = r^2, but in the Unit Circle r always equaled 1 and in order to make this true and correct, we would divide by r^2 on both sides, leaving us with (x/r)^2 + (y/r)^2 = 1. But wait...(x/r) and (y/r) look similar...wait....isn't that cosine and sine? Why yes, yes it is. But they're squared. Because they are squared, we have resulted towards a Pythagorean Identity. Which cannot be "powered up"or "powered down", because it is a Pythagorean Identity it must always be squared and no power greater and no power less. Also because the Pythagorean Theorem is a proven fact and formula that is always true, it is called an identity. We can prove this by demonstrating one of the "Magic 3" ordered pair from the Unit Circle (30*, 45*, 60*). We'll use 60* now theta of 60* is (1/2, radical 3/2) with 1/2 being x and radical 3/2 being y. Since we want to prove our derivation of x^2 + y^2 = r^2 we're going to use cos^2 +sin^2 = 1, so (1/2)^2 + (radical 3/2) ^2 =1.

2.
In order to derive the remaining two Pythogorean Identities we must divide the whole thing by cos^2x to find the tangent derivation which will lead us to tan^2x + 1 = sec^2x

And then we divide the original by sin^2x to find the cotangent derivation leading us to 1 + cot^2x = csc^2x. 

Now you might be wondering how did we know some of this information? Well this information was found from our Unit Q SSS packet, information we should have, say it with me, MEMORIZED!!!






Inquiry Activity Reflection

1. The connections that I see between Units N, O, P, and Q so far are...

  1. The most obvious one would be the role of sine and cosine in this Unit Q from the Unit Circle and its properties which can be seen in the reciprocal identities which are similar to the inverse of these trig. functions from Unit O and P. 
  2. In Unit P Concept 3 when using the distance formula for Law of Cosines with SSS or SAS, one of the formulas is a^2 = c^2(sin^2A + cos^2A) - 2bcCosA + b^2 and we know that cos^2A + sin^2A = 1. 

2. If I had to describe trignometry in THREE words, they would be... 

mentally-draining (this counts as one word), stressful, and insightful. 

Tuesday, March 18, 2014

WPP #13-14: Applications with Law of Sines and Law of Cosines

This WPP was made in collaboration with Sandibel Ramirez. Please visit the other spectacular, non-cheesebuckety posts on their blog by going here!

Well Phil has become a lonesome person with no friends after having invested so much time on his record store. BUT then his great friend Scott from freshman years comes back home to San Francisco and they quickly reconnect through spontaneous adventures through great 'ole San Fran.



a) Law of Sines


Problem: Scott and Phil agreed to meet at the record store so Phil can show Scott what he's been up to in life. They realize they're 15 feet apart. Scott heads to the store at an angle of N25E while Phil is heading off at an angle of N55W. Considering that Phil is a slowpoke, how far is he from the record store?


Solution:















b) Law of Cosines


Problem: Phil and Scott decide to go on a boat ride to relax and just ride the waves out after such a long day. Phil gets to the harbor before Scott however and told him to amount the SS Kirch at 5pm. At 5pm, Scott gets aboard the SS Cheesbucket. Phil's boat, the SS Kirch, is traveling at 80 mph and is at a bearing of 220* while Scott's boat is traveling at 115 mph at a bearing of 335*. It takes four hours for both "men" to realize they aren't aboard the same boat (they're huge ships and cell reception isn't exactly great out in the waters.) How far apart are the boats when Phil and Scott finally realize they're aboard different ships?


Solution:


 Remember that bearing always starts from the top, so once we draw both angles according to our bearings, we extend the lines and connect them with a line we'll call 'x'. In order to find our angle we use what we have, if the bearing is 220* and we subtract 180* we're left with 40* and if we subtract 335* from 360* we are left with 115* as our angle. 
Now Scott's boat, the SS Cheesebucket is traveling at 115mph for 4 hours then it has traveled a distance of 460 miles. Phil's boat, SS Kirch travels at 80 mph for 4 hours so really has traveled 320 miles. 
Our triangle is SAS, making sure our included angle is between two sides. Now that we have this information we can plug it into our equation of Law of Cosines, a^2 = b^2 + c^2 - 2bcCosA, which we plug straight into our calculator and then square root to find that the distance between Scott and Phil is a tragic 662.13 miles apart...someone better start swimming now.

Sunday, March 16, 2014

BQ #1: Unit P: Concept 2: Law of Sines SSA and Concept 4: Area of An Oblique Triangle

2. Law of Sines


In an SSA triangle, our three angles are not carved in stone as AAS or ASA triangles where even though we had two angles, in reality we had all three because all one had to do was add the two and subtract it from 180 to get the third angle's values, meaning there was no ambiguity there.
With SSA we only know one angle's value! (and two sides) There is a possibility however that with only this information that there may be no triangle at all! We always start the problem assuming there is two possible triangles, until you "hit a wall" which will lead to only one possible triangle or no possible triangle. A "wall" can either be if SinA=1.1 or 2.8, something that is not possible. Remember from our unit circle trig functions, we know that sin and cos are not compatible and do not work with numbers greater than one, (-1 less than or equal to sin theta less than or equal to 1), or if the angles add up to be something that is greater than 180.
In this triangle, we are solving for angle B and C, and side b. We have information for both angle A and side a, our magic pair, and we have information for side C but not side c, which is our bridge. We set our magic pair and bridge in proportion to one another to solve for the angle of C and use inverse sin to find it. But because we are assuming we have two triangles, and we are using sin (remember that with sin, we have positive answers in the first and second quadrant but our calculator only gives us the positive answer in the first quadrant), we use the reference angle of the first quadrant and subtract it from 180 to find the measurement of angle C'. In this case the angle of C' of 171 added with angle A of 155 was much bigger than 180 meaning only one triangle is possible. Had it been an acute angle like angle C of 9, there  could have been a possibility of two triangles.

4. Area Formulas

In a normal triangle where the area could be found with the formula of A = 1/2bh, we know that the perpendicular height of the triangle is h and b is the base. While the oblique triangles (where all sides are different lengths) area is one half of the product of two sides and the sine of their included angle <- very important to realize, it must be in between the two sides. 
We know that sinC=h/a from concept 1, so once you mulitply 'a' on both sides h (height) = asinC, and plugging 'h' into that A=1/2bh formula we really have A=1/2b(asinC)

Wednesday, March 5, 2014

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

Our dear friend Phil has been struck with the wish to travel around the country, he's been bitten by the wander bug.

The Problems


a) Phil packs his bags and takes a roadtrip up to Seattle. There he finds Seattle's Space Needle of 605 feet casting a 410 foot shadow at which Phil is standing at the end of. If Phil looks at the top of the building (and avoids getting his eyes burnt by the sun) what is the angle of Phil's eyes to the top of the building? (to the nearest hundredth of a degree AND assume Phil's eyes are 5 feet above ground level since he's a short little fellow.(http://upload.wikimedia.org/wikipedia/commons/3/38/BMX_aloft_and_Space_Needle_03.jpg)

b) Once Phil gets tired of all the Space Needle this and Space Needle that, and the city overall in general, he heads over to Yosemite where Moro Rock lies standing tall at 300 feet where he can enjoy the panoramic view. But the mighty rock is slanted at its highest point, Phil estimates the angle of depression from where he is standing (at 300 feet) to the bone-crushing  bottom to be 53*. Should Phil fall, like the klutz he is, how long is his path to certain doom? (Round to the nearest foot.) 


http://triggerpit.com/wp-content/uploads/2012/01/sequoia-national-park-moro-rock.jpg

The Solutions 


 a) Because it is from Phil's eye level of 5 feet we subtract 5 from 605 giving us 600. Its height is now 600 and the length is 410 which gives us the opposite (600) and adj (410) side. So we are looking for the angle (x). We use tan because of TOA and we use the inverse to undo tan to find our angle so it is tan -1 x (600/410) which is 55.65 degrees.


b) We know that the height is 300ft and the angle of depression to be 53*. We have the opposite side of 300 and we are looking for x which is the hypotenuse. Opposite and hypotenuse, SOH - Sin. So  this time it's sine of 53. So sin 53 = 300/x, we multiply by x on both sides which leads to us dividing 300 by sin 53 which equals approximately 758 feet. 


Monday, March 3, 2014

I/D #2: Unit O - How can we derive the patterns for our special rights triangle?

Inquiry Activity Summary 

1. 30-60-90 Triangle


First of all, make sure you recognize that this is an equilateral triangle, whose angles sum is 180*
We cut straight down the triangle because if 180* is the total sum of the triangle's angle we are left with 30* at the top, 60* at the side, and 90* at the bottom of the triangle from 30*.
And since it is an equilateral triangle, all of the sides are of equal length of 1. But since we cut the triangle in half, the bottom side's length is cut to 1/2. With this information, we're able to find the height since the hypotenuse (c) equals 1 is already given to us and we know that one of the values is 1/2, let's call 1/2 (a) for now. If we plug these values into the Pythagorean Theorem of a^2 + b^2 = c^2, **refer to the first picture for clarification** we realize our height is radical 3/ 2. 
If we give 1 a variable instead, like n, we are left with 30* - 1/2n, 60* -  radical 3/ 2, and 90* - 1n. But we use the variable 2n so our general pattern doesn't have fractions. By doing this, we have the pattern we all know, love, and recognize of 30* - n, 60* - n radical 3, 90* - 2n. The ratio of n : 2n is to show the relationships between the sides (and like I mentioned, using the variable 2n only makes deriving easier since it takes those pesty fractions away); it only expands as the ratio remains; such as 2: 4, 3 : 6, 4: 8, etc.

2. 45-45-90 Triangle


 This time we are given an equilateral square whose sides' lengths are 1. Now a squares' total angle sum is 360* meaning each corner has an angle of 90*. We cut diagonally in half because by doing so, we cut 90* corner angle into 2 which gives 45* angles - exactly what we want! We have two sides that equal 1 and we are looking for hypotenuse's value (which is that line by which we cut the square diagonally half by.)  In order to find our hypotenuse, we shall use the Pythagorean Theorem  of a^2 + b^2 = c^2. Our two equal sides of 1 shall be a and b. **look over photo for clarification** Now by inputting 1 for a and b, our c value equals radical 2.
To find our pattern, we use a variable - which represents any number. Let's use 'n' for now. By using n to represent our two equal sides of 1, our hypotenuse now equals n radical 2; the variable does allow for the relationship to stay consistent and allows us to expand it should the case ever be that the side length equal a number greater than one (this also applies to the 30-60-90 triangle.)

Inquiry Activity Reflection

Something I never noticed before about right triangles is how we are able to find these two triangles in equilateral shapes which just allows the whole derivation to be much easier; but something that I had forgotten and was brought back to my attention was the total angle sum of these equilateral shapes that allows us to create the triangles in the first place by cutting these shapes up either diagonally or straight down the half.  
Being able to derive these triangles myself aids in my learning because heaven forbid that I should ever have a brain fart on these concepts, I'll be able to fully comprehend special rights triangle and remember how to solve problems such as concept 7-8 by going back to these basics. And also being able to understand why a 45* side value is n and why a 60* side's value is n radical 3 and not because of that cute little memory trick of the 60's being radical times, which can also be considered with the 45-45-90 triangle with the 90* side value of n radical 2 because the 90's were also radical times.