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.