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

[cdn.screenrant.com/wp-content/uploads/He-Man-Masters-of-the-Universe-Movie.jpeg]

He-Man and his comrades are heading towards the battlefield, in a planned face-off against Skeletor and his goons. When they confronted their enemies, they had found an unfortunate situation: Skeletor's forces were on higher ground than them (on a mountain), leading to a widely unfair advantage. He-Man thought out the situation while his comrades kept the enemy busy, and had decided to calculate whether or not throwing a projectile at Skeletor was practical or not. He had estimated about 200 feet of distance between them (from He-Man to the base of the mountain), and a 70° angle of elevation to Skeletor. How far would his throw have to be, in order for his projectile to (diagonally and in a straight line) reach Skeletor?

  S(keletor)

   |\

   |  \

   |    \                                  cos 70° = 200 ft.

   |      \                                                    h

   |        \

   |          \                                  h x cos 70° = 200 ft.

   |            \

   |              \                                     200 ft.

   |                \                                  --------- = 584.76 ft.

   |                  \                                cos 70°

   |             70° \ H(e-Man)

         200 ft.

Seeing as how a throw that would have to travel 584.76 ft. was impractical, He-Man had simply decided to find a nearby mountain with a higher elevation, in order to reduce the required power in that throw. After a short while, He-Man finds a fitting spot upon a mountain, measuring 600 ft. in height. If Skeletor's elevation was 549.5 ft., and the angle of depression was 10°, then what would be the distance that He-Man's new-and-improved throw have to be?

I/D #2: Unit O - How Can We Derive the Patterns For Our Special Right Triangles?

1) We may derive a 30/60/90 degree triangle's special pattern from an equilateral triangle with a side length of 1, by first dividing the triangle into two halves that are completely equal. You'd split it right in the middle, creating two triangles that share one side of the previously whole equilateral triangle (which equaled 1), and the side that was created by splitting them in half. They each have their own side of 1, across from the 90° angle that they both have, which was created by the division.

Since we divided the side which had previously equaled 1 into two equal parts, then each triangle must have a side of 1/2. So, if we proceed with that deduction, then that side, squared, added with the other side that's not the hypotenuse, squared, should equal the hypotenuse, squared. Basically, a² + b² = c² (Pythagorean Theorem), with 1/2 being a, the other leg being b, and 1 being c. This works because the triangles we had created, have 90° angles, and therefore, have an assumed hypotenuse across from them, along with the values for a and b that fit into the aforementioned equation. (1/2)² + b² = (1)² would be 1/4 + b² = 1, and subtracting 1/4 from both sides would result in 3/4 as the value for b². And, if we take the square root of the entire fraction, then it should equal √3/√4, which would reduce into √3/2.

Lastly, we'd have to save this extremely rare knowledge, by multiplying everything by 2, and then, by a variable, so that this pattern will be easily transferable to any 30/60/90 degree triangle. Multiplying each side by 2n, basically, would yield a hypotenuse of 2n, a value of n for the side across from the 30° angle, and a value of n√3 for the side across from the 60° angle. This pattern shall forever exist for any 30/60/90 degree triangle, since the pattern was derived from ratios that will be the same for all 30/60/90 degree triangles.

2) We may derive a 45/45/90 degree triangle's special pattern from a square with the equal sides of 1, by diagonally dividing two 90° angles (across from each other) exactly in half, and making four 45° angles - two for each triangle. Each triangle would then, each have two sides that equal 1, and share a hypotenuse.

So, since the two triangles that we have given life to have 90° angles, then the Pythagorean Theorem can be utilized in order to find the length of the hypotenuse in this set-up. 1² + 1² = c² shall become c² = 2, and from that point onward, be √2.

In order to again, preserve our newly-discovered and revolutionary (in the field of mathematics) pattern, we must multiply all of our triangles' values by n, in order to create a distinct pattern that we can follow for any 45/45/90° triangle. Everything just basically gets an n attached to it.

Inquiry Activity Reflection
1) Something I never noticed before about special right triangles is that their established patterns can be deduced from a logical procedure, instead of just being "there", because we were told that it happened to be "there".

2) Being able to derive these patterns myself aids in my learning because I'm not just remembering given facts or something - I'm actually finding out this stuff myself.

I/D #1: Unit N Concept 7 - How Do Special Right Triangles and the Unit Circle Relate?

Inquiry Activity Summary
1) The first and foremost part of interpreting a 30° special right triangle, is setting the side opposite of the 30° angle to x, and going along with that, the hypotenuse as 2x, and the other leg of the triangle as x√3. In order to align this set-up with the unit circle, we must set the hypotenuse equal to 1, since the radius of a unit circle will always be 1, and the hypotenuse will be exactly between the Unit Circle's center and its arc. So, if we set the hypotenuse to 1, then we basically divide all of our previously-acquired values by 2x, making it proportionate to the 2x = 1 ratio that we had fabricated. So, the x would become 1/2, and the x√3 would become √3/2.

2) In order to make sense of a 45° special right triangle, we should first set up the hypotenuse as x√2, and the other two sides as x (since both sides are the identical, according to geometry). To match this up with the unit circle, we set the hypotenuse to 1, which is most simply done, by dividing x√2 by x√2, which gives us 1. Dividing x by x√2 would make x times (1 / x√2), creating a situation in which the x's cancel out, and you are left with 1/√2. In order to rationalize that, you would multiply it by (√2 / √2), which would result in (√2 / 2) for your two other sides.

3) Lastly, the 60° special right triangle should include the exact same values as the 30° special right triangle would. All that one would have to do, is ensure that values accordingly match up with degrees. But, I suppose that I have enough time to explain this. Basically, the side opposite of your 60° angle would be set to x√3, your hypotenuse set to 2x, and your other leg, set to x. In order to substantiate a hypotenuse equal to 1, you would divide all of the values by 2x, and end up with 1 for the hypotenuse, √3/2 for the side that had previously equaled x√3, and 1/2 for the side that had previously been x.

4) How Does This Activity Help You To Derive The Unit Circle?
I suppose that this activity helped me more easily identify all of the values that encompass various common degrees in a unit circle, by listing out all of the possible values that I should know. If I were tasked to find out the coordinates of wherever any 30° or 60° angle's terminal side met with the Unit Circle's arc, there could only ever be three values that I need to know (their signs and or order depend upon which quadrant of the circle they are in): 1, √3/2, and 1/2. For the 45° angle, literally the only two values I would ever need to figure out coordinates of wherever the 45° angle's terminal side met with the unit circle's arc would be 1 and √2 / 2. This stuff is really useful in trigonometry and all.

5) The triangles drawn in this activity were in Quadrant I, ergo, their positive coordinates. The signs would change in differing quadrants, with Quadrant II implying a negative x-coordinate, Quadrant III, a negative x-coordinate and y-coordinate, and Quadrant IV, a negative y-coordinate.

Inquiry Activity Reflection
1) The coolest thing I learned from this activity was that from remembering three simple values (√3/2, 1/2, and √2 / 2), you can get everything that you need in the Unit Circle. All you'd have to do is remember what different quadrants do to the signs of the values, and all that other stuff.

2) This activity will help me in this unit because I can now, more easily remember the Unit Circle.

3) Something I never realized before special right triangles and the unit circle is that just by identifying which side is the longest of the two of a 30° or 60° angle, you can determine which has the √3/2, and which is the 1/2 (√3/2 is bigger than 1/2).

RWA #1: Unit M Concepts 4-6 - Conic Sections In Real Life

Parabola

1) Definition: The set of all points equidistant from a given point, known as the focus and a line, known as the directrix.

2) Algebraically: The algebraic equation that is used to standardize parabolas is as follows: (x-h)^2 = 4p(y-k) or (y-k)^2 = 4p(x-h). Basically, one would modify any given equation that is determined to be one of a parabola into this format, most usually, through completing the square. In order to distinguish equations of parabolas from equations from other conic sections, you basically need to know only one thing: one variable is square, while the other isn't. This is a distinct feature of the parabola that we don't find occurring within equations for circles, ellipses, or hyperbolas. Anyways, by looking at the value of p, and at which variable is squared, you may determine what the parabola would look like graphically, as in its orientation and all. 

Graphically: The first and foremost characteristic that one would find while examining the parabola, is that it looks awfully similar to the letter 'U'. Its orientation can be determined through the sign that is complementary with the value of p, and the variable that holds the square. If x is squared, then the parabola will either face up or down - up if the p is positive, and down if it is negative. If y is squared, then the parabola will either face right or left - right if the p is positive, and left if the p is negative. The point at which the parabola can be directly divided in half - that is, also considered its vertex - will be known as the vertex, and the line which divides it exactly in half, known as the line of symmetry. The focus point, which will be on the line of symmetry the distance of 'p' away from the center, will help one decide what the orientation of the graph is like - by way of being in the direction of the lines. Lastly, the focus point may also determine whether the parabola is fat or skinny - if it is close to the vertex, then it will be skinny; if it is far away from the vertex, it will be fat. AND the directrix is basically the opposite of the focus point, and is a line.

 
http://www.mathsisfun.com/geometry/images/parabola-formula.gif

Features Of: The vertex is most basically, (h,k), the focus, (h+-p,k) or (h,k+-p), depending on which variable has the square and what the value of p is, the line of symmetry, either h or k, again, depending on which value has the square, and p is the variable that precedes the variable without the square, divided by 4. Ah, and the directrix is the exact same thing as the focus point - save that it is determined by subtracting p if the focus is found through addition, and vice-versa - and, it is a line.

3) Real World Application: As demonstrated in the video below, there are many instances in real life in which parabolas occur. For example, when the teenager shoots the basketball into the basketball hoop, a parabola is seen - and even outlined, too. If it were practical, I suppose that basketball players would study up a bit more in math, in order to improve their shots.



4) Works Cited: http://mathworld.wolfram.com/Parabola.html
http://www.mathsisfun.com/geometry/images/parabola-formula.gif
https://www.youtube.com/watch?v=FXuTKpp-vME

WPP #10: Unit L Concepts 9-14 - Probability

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WPP #9: Unit L Concepts 4-8 - FCP, Combinations, and Permutations

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SP #6: Unit K Concept 10 - Writing a Repeating Decimal as a Rational Number Using Geometric Series

        Concerning this problem, one in which you are tasked with finding the fraction of a number with repeating decimals, there are not really many things one has to remember to do. Though, it is a bit important that one remember to add whatever you didn't account for in your geometric series. In my case, it was the 4, since it was not part of the repeating numbers. You must also remember to plug your answer back in, to see whether or not you did it correctly. I mean, you don't really have to, but it is quite useful for saving a bit of face when having your problem examined by your colleagues. Yep.