Problem: Julian decided to take a boat ride on his brand new yacht. He takes it out for a spin and notices a cliff upcoming ahead. He wonders, if the angle of elevation to the top of the cliff is 52.53 degrees, and the base of the cliff is 1236 feet away from the boat, how high is the cliff? (Remember round to the nearest foot)
Ivan was on top of a lighthouse observing Julian on his brand new yacht. He looked around and noticed at the bottom of the cliff where he was positioned, that there was a boiling crab. He found a parachute and thought, " I wonder how long will be the path I glide if the angle of depression is thirty degrees?" Keep in mind he is already 300 feet above ground.
For this activity we had to derive the special right triangles not from the unit circle though. We had to derive the two special kinds ,45-45-90 and 30-60-90 right triangles. Each are completely different we had to find out what n was and why n couldn't just be a number.
How can we derive the 45-45-90 triangle from an square with a side length of 1?
We know that if we are given a square with side lengths of 1 we will have two right triangles if we split the square in half from the corners. For a 45-45-90 triangle we know the base and height are the same,1. If we do the Pythagorean theorem we will see that the hypotenuse will equal radical 2. N is there to be any given value, without n the triangle sides cannot be altered to match the initial constants, it is multiplied by the initial constants. For example if n were to equal 2 then the base and height are 2. The hypotenuse will equal 2 radical 2.
How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?
To derive the pattern for the 30-60-90 triangles we have to cut an equilateral triangle in half straight down the middle. Each side length of the triangle is one so when we split the triangle in half the base turns into 1/2. We also split the triangle in half to get 30 degrees as one of the angles. Then we will have a 30-60-90 triangle. Then we notice that we do not have the height of the triangle, we have to use the Pythagorean Theorem. Once you get the constants they can be altered to get rid of the ugly fractions by multiplying by two ( look at picture). The constants are the same because they are all proportional and everything was multiplied equally.
Inquiry Activity Reflection:
Something I never noticed before about special right triangles is how we have to tweak equilateral triangles and squares to get the sides for n.
Being able to derive these patterns myself aids in my learning because now I can refer to this square or triangle if I ever forget what the sides of the triangle equal.
The 30* triangle has three different sides to it: adjacent which is x, opposite which is y, and hypotenuse which is r. The side opposite the hypotenuse will always be x.The hypotenuse must be one if you want to derive the unit circle from the triangle. If you want this to happen you have to divide each side by 2x. Once you do this you will get x=radical 3 divided by 2, y=1/2 and r=1. We can use these simplified values as coordinates to determine where 30 degrees lies in a quadrant on the unit circle. We know that 30 degrees on the unit circle is located on radical 3/2, 1/2. This can be used for 150 degrees, 210 degrees, and 330 degrees. The only difference is that they are located on different quadrants and there will be negatives and positives.
2. The 45* Triangle
The 45* triangle has two sides that are the same length which are x and y the hypotenuse is r. To derive the unit circle from the triangle we have to get the hypotenuse to equal 1. In order to do this we divide every side by x radical 2. Once we get one on the hypotenuse we can get r and figure out the points. For the angle of 45 degrees we plot radical two over two, radical two over two. This is where it will lie on the unit circle. This will also stand for 135, 225 and 315 degrees. The only differences are the quadrants the negatives.
3. The 60* Triangle
The 60 degree triangle is the same as the 30 degree triangle it has three different sides. In order to derive this triangle from the unit circle is to divide the hypotenuse to get one. This is the same as the 30 degree triangle because you divide by 2x. When you get your final answers you will be able to plot the points. The points for 60 degrees are 1/2, and radical 3/2. These rules also apply for 120,240, 300 degrees.
4.
This activity helps me derive the unit circle because the triangles reflect different points on the unit circle throughout all of the four quadrants. Each of these triangles can be found in all of these quadrants and are all the same the only exception is that there are negatives and positives and they are located in different quadrants too.
5.
The triangle in this activity lies in quadrant one both the x and y values are positive which means that it is in quadrant one.
Inquiry Activity Reflection: 1. The coolest thing I learned from this activity was how you can find the points on the unit circle by using the special triangles. 2. This activity will help me in this unit because it can help me memorize where different points are and where some points lie on different quadrants. 3.Something I have never realized before about the special right triangles and the unit circle are that both these are in relation to each other when the hypotenuse is equal to one
All there is you need to know about Hyperbolas 1. Mathematical Definition. A hyperbola is "a curve where the distances of any point from a fixed point 9 (the focus)and a fixed straight line (the directrix) are always in the same ratio."(mathisfun.com)
These two pictures show the two formulas used for a hyperbola. The reason there are two different formulas is because the way each hyperbola branches out. This means that if the equation begins with x it will open up on the x- axis. You will also know that it is the horizontal transverse axis. If the equation begins with y it will open on the y-axis and you will also know that it is the vertical transverse axis. The different parts of this conic section on this axis are vertices, foci, and center. The conjugate axis includes co-vertices. The conjugate axis is the direction in which the hyperbola opens. Asymptotes are also very important they are shown as y=mx+b. The formula to find the asymptotes for a transeverse axis which is horizontal us y=k+ or - b/a (x-h). The formula to find the asymptotes for a transverse axis which is vertical is y=k + or - a/b (x-h).
To find the center which is (h,k) they will be included in the formula so they are paired off with x and y. X is always with h and y is always with k. To find the vertices you use what you have already which is the center. For example if the hyperbola y is first then you find the center and then use the x from the center to find the vertices which is the number for a up and the number for a down. The same with the co-vertices except you go left and right. A and B are found when you take the square root of the denominators. If y is first a squared is underneath it and b squared is underneath x or vice versa. To find the asymptotes you use the formula given in the above paragraph.
This video explained how an ellipse and a hyperbola are similar even with their foci. The eccentricity of a hyperbola should be 1 or above.The closer the eccentricity is to one the curves of the hyperbola will make it look sharper or more pointy. The farther away the eccentricity is to one it will make the curves of the hyperbola look straighter, which means it will look fatter. A smaller foci means a small width and a larger foci means a larger width.
Hyperbolas can be found all over the real world for example a common household lamp can cast a shadow that is a hyperbola. Also in architecture like the Dulles Airport which is shaped like a hyperbola. The one i picked is a nuclear power plant's cooling tower. The reason that this is the standard shape for all cooling towers is because it needs to withstand high winds and be built with little material. When the cooling tower is in the shape of a hyperbola it is easier for the building to take the damage of high winds. This shape also is very inexpensive to build with steel beams and concrete.
As you can see the cooling tower looks like a cylinder. Well when you take a cylinder by each side and push one side forward and one side back you will create a hyperbola. Hyperbolas can also be found with jets. As a jet breaks the sound barrier it releases a sonic boom. When the boom is released a cloud is formed and it looks like a cone, when it comes into contact with the ground then it becomes a hyperbola.
4. References
Vertical and Horizontal Transverse Axis of Hyperbola: http://www.mathwarehouse.com/hyperbola/graph-equation-of-a-hyperbola.php http://www.mathwarehouse.com/hyperbola/graph-equation-of-a-hyperbola.php
This is my student problem number six and it shows how to find partial sums of arithmetic series. One thing to always remember is to make the arithmetic series you have to separate the numbers so that they turn into a series. Also, remember to plug everything into your calculator the right way so that you won't get the wrong answer. I hope you enjoy the problem and make sure to check your answers. Thank you.
