BQ# 7: Unit V

     The origins of the the difference quotient comes from a graph and using an old equation from early this year.

Here is the graph f(x) and the line that's barely touching it at x is the tangent line. So the coordinates for the graph is ( x, f(x) ). 

This is another graph that has a secant line going through it. It still has the original point as the first one but it has another point in it. Since we moved a little bit from the original to the new point, then it's a change in the graph. That change can be written as delta x or as i put it h. So the new coordinates for this graph is ( x, f(x) ), ( x+h, f(x+h) ).

Highlighted in blue is the two coordinates from the graph. Highlighted in green is the slope formula that will help find the difference quotient. The one in pink has everything plugged in and in the denominator we see that the xs' cancel so there's only an  h left in the denominator. The last one in purple is the difference quotient and that is how we get the equation.

BQ #6 : Unit U Concepts 1-8

1) A continuity graph is a graph that's predictable, it has not breaks, holes, and jumps. Also the limit and the value are the same, the limit is the intended height while the value is the actual height. There are two groups, the removable and non-removable, a continuous graph is in the removable group.

A continuous graph is in the removable group because it has a limit. This can be considered a continuous graph because it has no breaks, hole, or jumps.

 Discontinuity graphs are in the non-removable group because these graphs have no limits. These graphs have jumps, breaks, and not predictable at one point. There are three types of discontinuity graphs, jump discontinuity, oscillating behavior, and infinite discontinuity.

Highlighted in green would be our jump discontinuity, orange is oscillating behavior, and in blue is infinite discontinuity. None of these graphs have a limit but only jump discontinuity can have a value. The other two have no limit or value. 

2) In this unit a limit is the intended height of a function. It only exits in a point discontinuity graph.

This graph is a point discontinuity and it has a limit. Now the first one is continuous but is still has a limit, the second one has a hole but it still has a limit because it's the intended height.

All of these don't have a limit so the limit does not exist for them.

Just like in the first question the difference between a limit and a value is that the limit is the intended height while the value is the actual height.

3)

To evaluate a limit numerically we do it by using a table that we can plug in to our graphing calculator. We can find the limit using the graphing calculator by tracing the graph. Then we can tell if the limit was reached.

 

When it's done graphically we can either use our fingers to see where the limit is if we come to the middle from the left and right. Then we just look at what type of discontinuity it is and if the limit exists or not. 

A

B

C

When it's done algebraically we can use three methods but the one we will always try first is the substitution method, know as picture A. Ifa problem can't be done using that method then we use the dividing/ factoring method. If those two don't work then we use the rationalizing/ conjugate method.

BQ#4: Unit T Concepts 1-3

Why is a "normal" tangent graph uphill, but a "normal" tangent graph downhill? Use unit circle ratios to explain.

    The main reason why their graphs are different is because of their asymptotes and their location. The graphs are based on their Unit Circle ratios, and the graphs can't touch the asymptotes. The asymptotes themselves are based on the Unit Circle.

Both tangent and cotangent have their asymptotes based on their Unit Circle ratios. The ratio for tangent is y/x and in order to how and asymptote x has to 0 so it can be undefined. If we look at the graph, we see that 90 degrees and 270 degrees has the ratio to be undefined. 90 and 270 degrees in radians are pi/2 and 3pi/2, and those are the asymptotes. The same thing for cotangent except that the ratios are switched, it's x/y. So in order for cotangent to have asymptotes, y must be 0 and 180 and 360 degrees have those points. The asymptotes for cotangent, in radians, is pi and 2pi.

 Tangent:

If we look at it already graphed we can see that 2pi and 3pi/2 are the asymptotes and the graphs can't touch them. Also, based on the Unit Circle, the four quadrants and if they are positive and negative. Both quadrants 1 and 3 for tangent is positive so it's above the x-axis. Quadrants 2 and 4 are negative so the graph is below the x-axis.

Cotangent:

The asymptotes for cotangent is pi and 2pi and just like tangent, the graphs can't touch them. The Unit Circle and it's four quadrants are just like tangent, 1 and 3 is positive and 2 and 4 are negative. Expect that the graphs of cotangent will not be the same as tangent because the asymptotes make the graph go downhill to follow the positive- negative of the quadrants.

BQ# 3: Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each trig graph?

Sine will be in Green.
Cosine will be in Orange.
Tangent will be in Blue.
Cotangent will be Yellow.
Cosecant will be in Pink.
Secant will be in Purple.

Tangent:

To see where the tangent graph will be we must first know about it's ratio. Tangent's ratio is y/x or sin/cos, cosine will determine where tangent will be. Where sine and cosine are both positive or negative then tangent will just be positive. If one of them is negative then tangent is negative. Tangent's asymptotes is determined by cosine, if cosine is 0 on the x-axis then that is one of tangent's asymptote. 

 Cotangent:

It's almost the same thing for cotangent except that it's sine that needs to be 0. If both sine and cosine are positive or negative then cotangent will be positive. If one of them is negative then cotangent is negative. The reason why cotangent is downhill is because of the asymptotes and where they are found. Since it has to be positive in quadrant 1, the cotangent graph has to be above the x-axis. Same thing for quadrant 2 but it's negative so it's below the x-axis.

Cosecant:

Where sine is 0, the asymptotes for secant will be those points. If sine is positive in two quadrants then cosecant will be positive, if it's negative, like in quadrants 3 and 4, then it's negative. Cosecant relates to the graphs by it's shape, the shape is between two asymptotes. So the asymptotes basically determine the shape and the asymptotes is where sine is 0.

 Secant:

Where cosine is 0 the asymptotes of secant can be found. The way the secant graph looks like, positive or negative, is the same as the cosine graph. The shape  of is is also determined by the location of the asymptotes and the cosine graph.

BQ# 5: Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.  
       Sine and cosine do not asymptotes because asymptotes occur when the denominator of the ratios is 0 (undefined). The other trig graphs can be divided by 0 and it can be undefined while sine and cosine can't be divided by 0. They can only be divided by 1 because 1 is their restriction on the Unit Circle and on the Unit Circle itself it goes from 1 to -1 on both axis. They both can't be divided by 0 because it's not undefined but no solution. The other four especially tangent and cotangent have no restrictions and if their ratios equal undefined then it's their asymptote.

BQ#2: Unit T Concept Intro

A)   When is comes down to sine and cosine, their periods are are 2pi and it's like that due to their similarity to the Unit Circle.

Sine in the Unit circle is positive in the first and second quadrant and negative in the third and fourth quadrant. If we start from zero and do one complete rotation to get it to be positive again, then then it was a 360 degree or 2pi rotation. So when we stretch Unit Circle into a line, the cyclical will be stretched out too because the first two quadrants will be above the x-axis and the last two will be below the x-axis. 

The same thing goes for cosine since quadrant I and IV are positive and quadrant II and III are negative. We need to start at 0 degrees and as we move around in the Unit Circle well reach the positive when we get to 360 degrees. When it reaches 360 degrees the it made one complete rotation. If the stretch out the Unit Circle to a straight line for cosine then we will start above the x-axis since quad. I is positive and for quads II and III it will be below the x-axis. As we approach quad IV to make a cyclical then it go back up above the x-axis.

For tangent it's different because it's only half the trip to get a cyclical. Quadrants I is positive and quadrant II is a negative so we already have our cyclical. On the Unit Circle, it will start at 0 degrees and when we reach a positive again it will be at 180 degrees and the radian value is pi.

B)    Sine and cosine only have an amplitude of 1 because they have a restriction of 1 and -1. If we use any number and use it in their ratio, it will not work except those numbers less than 1. We only use the 1 in the Unit Circle then sine and cosine will work because it is in their restriction,  if it wasn't then it will be undefined and will not work.

Reflrction #1- Unit Q: Verifying Trig Identities

1. What it means is that when we are given a problem we try to use all three forms of identities to solve one side to see if it matches the other untouchable side. In other words we are verifying to see it we can get the same answer.
2. For me to solve them more easily, I try to get them to equal sine or cosine. I mostly try to see if any part of the problem can be made into a ratio or reciprocal identity and if they can be multiplied. If they are then it would be even better if they are written as fractions so I can cancel things out.
3. The first thing I try to see is if the problem is a simplifying or verifying and then I can try to use the conjugate if it's needed. It doesn't matter if there is because I can substitute in an identity or if it's a fraction combine them. If the problem comes out to have a monomial then it's even better because I can separate them and hopefully it turns to an identity.

SP#7: Unit Q Concept 2: Finding all trig. function

Please see my SP7, made in collaboration with Leo Escutia, 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

INQUIRY ACTIVITY SUMMARY:
     Pythagorean identities come from both the Unit Circle and the Pythagorean Theorem. First to clarify, an identity is "an equation that is true no matter what values are chosen". The Pythagorean Theorem is seen as an identity because no matter what two values you have and when you solve it it will give you the third value. If you check your work by using the found value and a value given, the answer will be that same given value you did not use. To get the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, I will work it out using the Unit Circle and the Pythagorean Theorem and I will show and explain it in the following pictures:

First I'll show it by using the Unit Circle using the first quadrant since sine, cosine, tangent, co-secant, secant, and cotangent are all positives. If we plot a triangle in the quadrant we can already see the connection between the two. Th height will be "y" since it follows the y-axis, "x" will be base since it's on the x-axis, and the hypotenuse will be "r" because the hypotenuse is the radius of the circle. When it gets plugged in to the Pythagorean Theorem "a" will be "y", "b" will be "x" and "r" will be "c".

When I use "x,y and r" in the Pythagorean Theorem it will make sense if it's written out like y^2 + x^2 = 6^2 because when we try to make it equal one then it will make sense. If I take the sine of angle A it will be y/r, the cosine of it will be x/r, and "r" will stay as "r". Now I can show how (y/r)^2 + (x/r)^2 = r^2 can equal one but there is a faster and easier way to get one. If I write out "opp" as opposite, "adj" for adjacent, and "hyp" for hypotenuse and then in Pythagorean form, it's opp^2 + adj^2 = hyp^2. If I divide both side by hyp^2 then the hyp^2 will be 1 and it will be opp^2/ hyp^2 + adj^2/ hyp^2 = 1. If I pug it in to the other equation I said, opp^2/ hyp^2 is the cosine of angle A and adj^2/ hyp^2 is the sine of angle A. So cos^2(theta) + sin^2(theta) = 1 is one of three Pythagorean identities.

To find the second one, which is the identity with secant and tangent, I will use cos^2(theta) + sin^2(theta) = 1 and divide both sides by cos^2. The cosines will divide to be 1, sin^2/cos^2 is the Ratio identity of tangent(theta), and 1/cos^2 is the Reciprocal identity of sec(theta). So we end up with 1 + tan^2(theta) = sec^2(theta).

    

To find the third Pythagorean identity that has cotangent and co-secant, it will be just like the second one except that I will divide it by sine^2. If I divide both side by sin^2 to both sides then cos^2/ sin^2 will be the Ratio identity of cotangent, sin^2/ sin^2 will divide to 1, and 1/ sin^2 is the Reciprocal identity of csc(theta).

INQUIRY ACTIVITY REFLECTION: 
1) The connections that I see between Unit N, O, P, and Q so far are the angles that can be found in the quadrants of the Unit Circle and how the triangles that are made in the quadrants can be used with the Law of Sine And the Law of Cosine to find any missing angle or side length.
2) If I had to describe trigonometry in THREE words, they will be hard, understandable, and progressive.

WPP #13 and 14: Unit P Concepts 6 & 7: Applications with Law of Sines & Law of Cosines

This WPP 13 & 14 was made in collaboration with Leo Escutia. Please visit the other awesome post on their blog by going here.
  

     One day Robin decided to go fishing but out at sea. He knew that his long time friend was fisher too and he had a boat. So his friend Gabriel took his to a good fishing spot off the coast. Hours passed but when they decided to head back, the engine failed. They radioed for help and two Coast Guard station responded back. They needed their exact location but the boat was acting weird and couldn't give the exact distance. Station B is due south of Station A and they are 100 miles apart. From what they can tell, the boat is S 45 degrees W of Station A, and N 65 degrees W of Station B. How far is each station to the boat?

Here we have it drawn to show how the triangle is suppose to look like. This a ASA problem since we give two angles and one side in between them.  Since we give two angles, the third missing side will be70 degrees.

We used the Law of Sine to solve for side "a". Angle C, which is where the boat is, will be our bridge to solve for side "a' and side "b". Sin 70/ 100 = Sin 45/ a, we cross multiply to give us a(sin 70) = 100(sin 450). We divide sin 70 to both sides to cancel sin 70 on one side so that it will be a = 100(sin 45)/ sin 70. Station A is 75.2 miles away.

Like I said, sin 70/ 100 is our bridge and it will be used to fine side "b". So sin 70/100 = sin 65/b, we cross multiply to give us b(sin 70) = 100(sin 65). We divide sin 70 to both sides so that sin 70 cancels on one side so that we can have b = 100(sin 65)/ sin 70. Station B is 96.4 miles away from the boat.

     Once they got rescued and taken back to Station A they are taken back home. They both leave the station at the same time, they diverge an angle of 100 degrees. If Robin is 4 miles away from the station and Gabriel is 5.5 miles away too, then how far away are they from each other?

With it's picture and work done together, we will use the Law of Cosine to find the missing side. It will be a^2 = 4^2 + 5.5^2 - 2(4)(5.5) cos 100. We just plug that in to our calculator and they are 7.3 miles apart.

BQ #1: Unit P Concepts 1 and 4: Law of Sines and Area of Obliques

Concept 1: LAW OF SINES
     We need the Law of Sines when are working on a triangle that is not a right triangle. When we have that we use the Law of Sines to solve it but only if the the triangle is an AAS or an ASA. How to derive the Law of Sines will be shown in these following pictures:

If we are given a triangle the A,B, and C as their angles and a, b, and c as their sides we can just split the triangle in half to form two triangles.

 

Now if we just focus on angle A, angle C, side a, side c and the height, h, I can show you how they can equal each other. If we take the sine of angle A, it will be h/c but since this is a part of a triangle we don't know what h is. So c will be multiplied to both side and it will give us h = c(sin of angle A). The same thing will happen if we do the sine of C, except it's h/a and we multiply a to both side and we'll get h= a(sine of angle C). We can get the height in two way but in the end the height will be the same for the both of them. Since c(sine of angle A) = a(sine of angle C), and we cross multiply, then ( sine of angle A)/a =  (sine of angle C/c.

If we stay with the same angles and side but with a different height then I can show how sine of B can be the same as the other two.

In this version of the triangle we'll use the sine of angle B and it is k/c but we don't know what k is so we multiply c to both side and we'll get k= c(sine of angle B ). On the other triangle it's the sine of angle C and that's k/ b. Just like the other one we multiply b to both side and get k= b(sine of angle C). But if  c(sine of angle B ) = k and b(sine of angle C)= k then they are both the same. Which mean that if we cross multiply them then (sine of angle B )/b = (sine of angle C)/c and if that is true then (sine of angle A)/a will be the same and all three will work and we'll get the same answer.

Concept 4: AREA OF OBLIQUES.
     The "area of obliques" is derived from the area of a triangle which is A = 1/2 bh. The base is b and the height is h in a right triangle. But if we don't have a right triangle then it will almost be the same but that depends on the side and angle given to us. If we have angle A, side b, and side c given to us then the area formula will be A = 1/2 b(c(sine of angle A). The formula can be rewritten to work with the two other angle. The following picture will show which formula will work for that problem.

WPP #12: Unit O Concept 10-Sovling Angles of Elevation and Depression Word Problems

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     A) Robin went on a trip to New York to visit his friend that he hasn't seen for a while. As days past, he's invited to a rooftop party in central part of New York. Before he enters the building he looks up at the top of the building. From ground level, he measures the angle of elevation to the top of the building to be 72 degrees. If Robin is 25 feet away from it, what is the height of the building?

     B) Robin is now at the party and having a good time until he looks at the taller building right across the street from the one he's on and sees his friend waving at him. He measures the angle of elevation to the building his friend is on to be 32 degrees and see the angle of depression to be 44 degrees. If the buildings are 60 feet apart, how tall is the building that his friend is on?

For part A we have one side already given to us and it's 25 feet along with one degree. We'll use tangent to help us find "x" and when we put it the way I have we have: tan 72= x/25. Well we multiply 25 to both side so 25 can cancel itself out on the right and we end up with: 25 X tan 72. We plug it in to our calculator and the height of the building will be 76.9 ft.   

Here we have both buildings to show the comparison and to find the height of the taller building, which is the one his friend is on, we need to do two parts to solve the entire height of the building. You can see where I marked to be all of "x" and you'll also see "A" for the side of the triangle with the angle of 32 degrees, the same for the second triangle with the 44 degree angle. 

For part B, we have two triangles like I stated in the picture above, the first triangle will use the 32 degree angle and use tangent to solve it. We multiply 60 to both sides and we have x = 60 X tan 32. We multiply 60 to tan 32 when we plug it in to the calculator and that missing side will be 37.5 ft. We do the same work for the second triangle but we use 44 degrees this time and for that missing side we get 57.9 ft. We add 37.5 ft to 57.9 ft and we get 95.4 ft to be the height of building Robin's friend is on.

I/D #2: Unit O- Derive the SRTs

INQUIRY ACTIVITY SUMMARY
     To get the patterns for a 45-45-90 triangle, it's better to use a square that has equal side of 1. From there we can split the square diagonally and get two 45-45-90 triangles. From there we can use the Pythagorean theorem to get the hypotenuse of the triangle and from there we can see the pattern starting to form. Now we use "n" to represent the ratio of each side and as a variable to represent any number that can take it's place. These next few pictures will show how we can derive the pattern of the 45-45-90 triangle.

THESE SET OF PICTURES ARE FOR A 45-45-90 TRIANGLE ONLY:

Here we have a perfect square with each side being 1 and each corner being 90 degrees. To get the 45-45-90 triangle we are going to do one step to get the triangle and the Pythagorean theorem to get the missing side.

To get our triangles we just split the square diagonally but we are just going to use the triangle highlighted in green. We already have two sides, the horizontal and the vertical side, and each side is 1. The triangle is still incomplete because we need the hypotenuse and to find it we'll use the Pythagorean theorem to find it.

It doesn't matter which side is a or b because both sides are the same. We plug it in the the Pythagorean theorem and we have 1^2 + 1^2 = c^2. We square the 1's and add them and we get 2 on that side. Now it's 2 = c^2, we square root each side making 2 into radical 2 and c^2 to c. So radical 2 will be our hypotenuse for the 45-45-90 triangle. We nearly done but we need to plug in "n" to each side, so the vertical and the horizontal sides will be "n" and "n-radical-2" for the hypotenuse side. 

 THESE SET OF PICTURES ARE FOR A 30-60-90 TRIANGLE.

Here we have an equilateral triangle with each side length of 1 and each angle being 60 degrees. To get a 30-60-90 triangle we split the triangle in half and get two of them but in this case we are only going to use the one highlighted green. The hypotenuse is already there so it's length is 1, the horizontal is also there but it's not 1 because we split it in half so it's actually 1/2. The third side will be found by using the Pythagorean theorem.

In this case, side "a" will be the horizontal side and side "b" will be the vertical side. When we plug it into the Pythagorean theorem, 1/2 will be squared and it will be 1/4 and 1^2 will just be 1. So we then subtract 1/4 to both sides and end up with b^2 = 3/4 but we need to square root both side to make b^2 to just b. When we square root 3/4 it applies to the top and the bottom, which means that 3 will be radical-3 and 4 will be 2 since the square root of 4 is 2. Our final answer for side"b" will be radical-3/4.  

This is how a 30-60-90 triangle is suppose to look, the hypotenuse side being 2n, side "a" being n, and side "b" being n-radical-3. Well when I left off in the other picture it was suppose to be radical-3 for side "b", 1/2 for side "a", and 1 for the hypotenuse. To get it to be the derived pattern we just multiply 2 to each side, so the hypotenuse will be 2, side "a" 1 because the 2's cancel each other, and radical-3 for side "b" because the 2's also cancel each other. The "n's" are put in to show that any number can take it's place.

INQUIRY ACTIVITY REFLECTION
 1. Something I never noticed before about special right triangles is how they are derived form other shapes like the square and the equilateral triangle.
2. Being able to derive these patterns myself aids in my learning because it shows that I know how to derive them and how they came to be a 45-45-90 and 30-60-90 triangle.

I/D #1: Unit N SRT and UC

INQUIRY ACTIVITY SUMMARY
    SRT really do connect with the UC, because they point out the ordered pairs for that specific angle and it tells us how we get that ordered pair. We have three ordered pair in a quadrant and it does not include the quadrant angle. For 30 degrees, 45 degrees, and 60 degrees they all have different ordered pairs. For the set of pictures that are going to be presented, they will be labeled as a SRT should be  and how to get it's ordered pair.

Hypotenuse or "r" will be in blue.
Horizontal side or "x" will be in pink.
Vertical side or "y" will be in green.

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 30 DEGREES ONLY.

Labeled as a SRT, "r" is 2x, "y" is x, and "x" is radical 3; but the hypotenuse but be 1.

 

Since "r" has to be 1, the only way to make it 1 is to divide 2x by itself. If we do that to one side, all the sides must be divided by 2x. As we divide, the x's will cancel for the "y"side, x is really 1x and it's divided by 2x so it will equal 1/2. For the "x" side, the x's cancel and we are just left with radical 3/ 2 and that will be our x value.

We have the triangle in it's new form with it's new values for "r", "y", and "x" and it's on a coordinate plane. The origin, which is 30 degrees, will be (0,0). As we find a point for the triangle on the coordinate plane, the hypotenuse and "y" intersect and that is where our order pair lies. Since radical 3/2 is "x"'s value and 1/2 for "y"'s value, that (x,y) point on the coordinate plane will be the same. So our ordered pair for a triangle with the origin of 30 degrees will be (radical 3/2, 1/2).

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 45 DEGREES ONLY. 

This is how a triangle with the origin point of 45 degrees looks like when it's labeled as a SRT. Just like the one before it, the hypotenuse has to be 1.

The hypotenuse of the triangle has to be 1 so what we do is multiply the reciprocal of x. X is not by itself, it's x/1 so when we multiply the reciprocal of it, which is 1/x, we'll get our 1. We'll have to multiply the 1/x to the other two sides. For the "y" side, it's x radical 2/ 2 and we multiply 1/x to it. The xs' cancel and we end up with radical 2/2 and that its value for the "y" side. The same goes for the "x", we do the exact same thing and radical 2/2 will be the value for "x" also.

This is the new version of the 45 degree triangle with its new side values. As we plot it to a coordinate plane, the 45 degree angle will be on the origin which is (0,0). With the ordered pair being (x,y) it will be just like the other angle before this one. The x value be radical 2/2 because it goes radical 2/2 to the right and radical 2/2 will be y because it goes up radical 2/2. So (radical 2/2, radical 2/2) will be our ordered pair.

THESE SET OF PICTURE IS FOR A TRIANGLE WITH AN ORIGIN POINT 60 DEGREES ONLY.

These triangles are just like the 30 degrees triangles that I did in the beginning, except that they have some parts switched around. This is how it looks like when it's labeled like a SRT, the hypotenuse is still 2x. The "y" side is now x radical 3 and the"x" side is just x.

Like the other two, "r" has to be 1 and we will have to divide 2x by itself to get 1. We will also have to divide the other two by 2x. For the "y" side, x radical 3 will be divided by 2x. The x's will cancel leaving us with just radical 3/2 and that is our answer for that side. The "x" side just has x and being divided by 2x the x's will just cancel and it will leave us with 1/2. X is not just x by itself, it's 1x so that's how we get 1/2.

Here it is on on a coordinate with 60 degree angle as the origin point. To get the (x,y) ordered pairs we will have to use the values of the "x" and "y" sides of the triangle. Since we go 1/2 to the right because that is value of the "x" side, 1/2 will be our x. Then we will go up, the value of the "y" side is radical 3/2 and that will be our y. So our ordered pair for this triangle will be (1/2, radical 3/2).

     From this activity and the explanation that I gave for each type of Special Right Triangles, helps derive each ordered pair that is on the Unit Circle. It shows how we get an ordered pair for a 30 degree, 60 degree, and 45 degree angle that is on the Unit Circle.
     For these types of angles and their ordered pairs, they are just in the first quadrant of the unit circle. Now the following pictures will show the changes of these triangles it the other three quadrants:

30 degree angles will be in blue.
45 degree angles will be in green.
60 degree angles will be in pink.

This is how all of the look like in the first quadrant, each degree with it's ordered pair. All of the ordered pair are all positive because they are on the positive side of the Unit Circle. Well in the Unit Circle we are just using a circle with a radius of 1. So x and y in quadrant 1 are both positive which means they are also positive.

 

In quadrant 2 I will just show the 30 degree angle and its ordered pair. In the second quadrant, x is now a negative. Which means that radical 3/2 is now a negative but the y is still a positive so 1/2 stays the same. This applies to the other two angles, what they have as x will be negative while the y stays positive.

This is in quadrant 3 and I'm using the 45 degree angle to show how its ordered pair changes from the first quadrant to the third. In the third quadrant both x and y are negative so the ordered pair of the 45 degree angle are negative. Both x and y on the coordinate plane are negative so every angle with an ordered pair in the third quadrant will be negative.

In the last and final quadrant, quadrant 4, I will use the 60 degree angle and its ordered pair to show the changes from quadrant 1 to quadrant 4. Well in this quadrant the x is positive and the y is negative. So 1/2 is x and it will stay positive while radical 3/2 is y, which means that it will be negative. This applies to the other two ordered pairs, their x's will be positive while their y's will be negative.

INQUIRY ACTIVITY REFLECTION 
1. The coolest thing I learned from this activity was the fact that the SRT tells us how these triangles get their ordered pairs and how change in each quadrant.


2. This activity will help me in this unit because it reveals where the ordered pairs are in each quadrant depending on their angle and which quadrant they lie in to determine if they are positive or negative.


3. Something I never realized before about special right triangles and the unit circle is that they both need each other to work. With out the SRT, there will be no ordered pairs in the UC and with out the UC the SRT will just apply to triangles and nothing else.