Unit Q Concept 2

Please see my SP7, made in collaboration with Vanessa Morales,by visiting her blog here. She also runs a cool blog and really you guys should go check her out.

I/D3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY

To find where sin^2x+cos^2x=1 we have to use the pythagorean theorem. X^2+Y^2=R^2. First we need to get R^2 to equal one. we do this by dividing R^2 by itself and divide everything else by R^2. we end up getting X^2/R^2+Y^2/R^2. After, we know that X^2/R^2= Cos^2x because we remember our unit circle. Y^2/R^2= Sin^2x. so we know that Cos^2x+Sin^2x=1.

It's good to know these, but these aren't the only trig functions we need to find. We also need the Tan^2x and the Cot^2x. We divide everything by Cos^2x. We come up with Sin^2x/Cos^2x+1=1/Cos^2x. Sin^2x/Cos^2x=Tan^2x since we remember the trig functions. The final answer will be Tan^2x+1=sec^2x.

To find Cot^2x we divide everything by Sin^2x this time.

We get 1+cos^2x+1/Sin^2x. from this we just simplify and we know that Cos/Sin = Cot because of our unit circle.  And 1/Sin is equal to Csc. so in the end we will get 1+Cot^2x=Csc^2x.

INQUIRY ACTIVITY REFLECTION

1. The connections that i see between Units N, O, P, and Q so far are the use of the trig functions throughout all of these. Also we have kept the unit circle through all of this.

2. If i had to describe trigonometry in THREE words, they would be eww (not sure if thats an acceptable word, if not then 'i don't like it'), difficult and helpful.

WPP #13 & 14: Unit P Concept 6 & 7 Law of Sines and Cosines.

This WPP13-14 was made in collaboration with Judith Quintanilla.  Please visit the other awesome posts on their blog by going Here .
Problems.

Law of cosines.

Billy Bob Joe Ray Eugene III decided to go quad racing with Camila. When they leave the lot where they rent the quads, Billy rode 70mph N52E. Camila rode at 80mph. and at an angle of 110 degrees bearing. suppose they rode for 3 hrs. and went in a straight line. How far would they be from each other?

Law of Sines

Camila and Billy Bob Joe Ray Eugene III want to race. They both see a red flag that is 240 ft from Camila and The angle formed at Camila is 54.14 degrees. The angle with Billy's is 58 degrees. How far is Billy from the flag? Is this fair?

Solutions

Cosines

In order to find how far they are we use the law of cosines. we plug x^2=240^2+210^2-2(240)(210)Cos58. and we get x=219.74 miles.

Sines

Sine we Know <A, Side a, and <B, we need to find side b. we get sin54.14/a=sin58/240. we manipulate it to get a by itself so it'll be a=240Sin54.14/Sin58 and we get a=229.36 ft. Billy side was shorter and he won the race!! WOOHOO!!

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

1. Law of Sines - Why do we need it?  How is it derived from what we already know?

The Law of Sines is used when we are trying to derive triangle that is not a right triangle.

We start with a non-right triangle. we drop an imaginary perpendicular line from <B and make two right triangles.

We call the height of the triangles h. Since they are now two right triangles, we can use the basic sig function. SOHCAHTOA. SinA= h/c. SinC=h/c. We use basic algebra to manipulate them.

Since both SinA/a=h and SinC/c=h, this means that SinA/a=SinC/c

4. Area formulas - How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?

The area of an oblique triangle os 1/2 of the product of 2 sides and the sin of their included angle.

Well we all know that the formula for a triangle is A=1/2bh where b is the base and h is the height.

the height is unknown so we use sinC=h/a.

We need to get "h" by itself so we just multiply the a to the other side so h=aSinc. We now know h so we plug it into the area formula (A=1/2baSinC)

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

Problems

a). Billy Bob Joe Ray Eugene III decided he take a break from his electric banjo to go visit the Empire State Building. When he was walking he saw it. From the ground, it's angle of elevation is 23.5 degrees. He knows that the building is 1250 feet tall. How far is he from the base building? 










http://upload.wikimedia.org/wikipedia/commons/c/c7/Empire_State_Building_from_the_Top_of_the_Rock.jpg





b). When he is on top of the Empire State he notices his favorite pizza place. He is hungry and wants to see how far it is. From the Building to the Pizza Place, the angle of depression is 35degrees. How far is the Pizza Place from the Empire state building?



http://haysvilleonline.com/hol/archives/625-The-Pizza-Place.html

Solutions

a). To find the how far Billy Bob Joe Ray Eugene III is from the building we have to use tan since we know the opposite and the angle. We need to find the adjacent so we use Tan23.5=1250/x because tangent is opposite/ adjacent. We manipulate it so we get x alone on one side so we get x=1250/tan23.5. We plug that in and get 2874.803 feet. So Billy Bob Joe Ray Eugene III still has some walking to do.


b).  To solve this one we have to use tangent because need to find the adjacent. We use Tan35=1250/x. We need to manipulate it to get the x alone. We should get x=1250/sin35.
when we plug it in and we get 1785.185 ft. This means our buddy here has to suck it up and eat a burger from cross the street.

I.D#2. :Unit O :Concept.7-8. How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY:

In order to derive the pattern for a 45-45-90 triangle from from a square with a side length of 1, we first need to cut the square from corner to corner, as shown in the picture, to get the triangle. Since the square has a side length of 1, we know that the x and y both also equal one so we can label the sides 1.

Now you have the sides, so all you need no is the hypotenuse. To find it we will use the pythagorean theorem. a^2+b^2=c^2. plug in the numbers we have. Since we know a and b both equal 1 then we use 1+1=c^2 since 1^2 still equals 1. We end up having C^2=2. so we have to cancel he square by finding the square root of c^2 which will equal C... But what you do to one side, you do to the other so it'll end up being the square root of 2.

Now that we have the  numbers, we will now add the variable because the sides are not always gonna equal 1. You can use any letter for the variable, i used X. Since both a and b both equal 1 we can just use X for both of those sides. To find the hypotenuse we have to use the pythagorean theorem. we plug in X^2+X^2= 2X^2

Now that we're done with the 45-45-90, we can move on to the 30-60-90 from an equilateral triangle with a side length of 1 and angles of 60-60-60. We start by cutting it right down the middle.

 Since we cut it right down the middle. the top corner is also cut in half so instead of a 60, its now 2 30's. And also the bottom is cut in half so instead of being 1 is it 2 .5's or 1/2's.

We, again, have to use Pythagorean theorem to find the the Y or b. First we have to manipulate a (1/2)^2+b^2=1. Once we do, it will end up being b^2=1-(1/4). That means b will equal to radical3/2.

since these are just for the side, the letter variables you use are the exact same thing. Hypotenuse=n

x=n/2 y=nradical3/2. To get rid of the fractions all you do is multiply each side by 2. Hyp.=2n  x=n  y=n radical3. N is just what you use when The hypotenuse is unknown. so the completed triangle should look like this.

INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles they are the exact same things as the Unit circle.

Being able to derive these patterns myself aids in my learning because it makes me independent and i don't have to rely on anything because i have all this in my head.