Tuesday, March 18, 2014
WPP #13 & 14: Unit P Concept 6 & 7: Applications with Law of Sines and Cosines
Please see my WPP13-14, made in collaboration with Adrie Garcia, by visiting their blog here. Also be sure to check out the other awesome posts on their blog.
Saturday, March 15, 2014
BQ #1: Unit P Concept 1 & 4 - Law of Sines and Area Formula
1. Law of Sines - Why do we need it? How is it derived from what we already know?
The law of sines is needed to figure out the missing parts (angles and sides) of a non-right triangle. Non-right triangles are more complex to solve for since they are not like special right triangles and have a special side relationships (ex: 45-45-90 triangle has a relationship of (x)-(x)-(x)radical 2).
The deviation to this law is shown through the process and pictures below:
First, there is the non-right triangle ABC. An imaginary perpendicular line is drawn to cut angle B. The imaginary line is labeled as h.
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 is derived from the triangle area equation and a trig function. The area of any triangle is (1a) A=1/2bh, where b is the base and h is the height. However, in an oblique triangle, the height is not given and the sin trig function must be used.
With the oblique triangle provided as an example, triangle ABC is cut perpendicularly in half from angle B to create 2 right triangles with the sharing side of h. Side h must be used with sin, in this example sin C is used with h/a (sin A could have been used as well). (1b)The sin trig function is rearranged to equal h.
The law of sines is needed to figure out the missing parts (angles and sides) of a non-right triangle. Non-right triangles are more complex to solve for since they are not like special right triangles and have a special side relationships (ex: 45-45-90 triangle has a relationship of (x)-(x)-(x)radical 2).
The deviation to this law is shown through the process and pictures below:
First, there is the non-right triangle ABC. An imaginary perpendicular line is drawn to cut angle B. The imaginary line is labeled as h.
The perpendicular line created (2) right triangles. From there, the basic trig functions of sin (opposite/hypotenuse) can be used for angle A and C. The trig functions are then rearranged to equal to h.
Since both functions are now equal to h, they can be equal to one another. Both are then over the corresponding sides used (sides a and c). Simplification can be done and the law is finished. (*Note: this implies the same for B if the perpendicular line was cut either from angle A or 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 is derived from the triangle area equation and a trig function. The area of any triangle is (1a) A=1/2bh, where b is the base and h is the height. However, in an oblique triangle, the height is not given and the sin trig function must be used.
With the oblique triangle provided as an example, triangle ABC is cut perpendicularly in half from angle B to create 2 right triangles with the sharing side of h. Side h must be used with sin, in this example sin C is used with h/a (sin A could have been used as well). (1b)The sin trig function is rearranged to equal h.
Since the sin trig function now equals to h, it can be substituted into the (1a) area of triangles. (*Note: the perpendicular line from the beginning can cut any angle and create in total 3 ways to find the area with the same concept).
It relates to the formula we are familiar with since it is mostly substitution, the new equation has parts from the old one. The triangle is still multiplied by 1/2 and consists of 2 sides. However in the new one, it also consists of an angle which is in between the two sides used. In the original one, meanwhile, uses the base and the perpendicular height.
Wednesday, March 5, 2014
WPP #12: Unit O Concept 10 - Angles of Elevation and Depression
Problems
- Farco decided to put a flag vertically on top of his ice-cream building. After having it set up, he wonders how high up is the the tip of the flag from the ground. He stands 5 feet away from the building and at ground level the measure of elevation is 76 degrees. a) How high up is the tip of the flag in the nearest foot?
- Farco decides to stand on top of the building, which is 15 ft high, and leans against the flag post. From that view, he sees an army of starving children on the other side of the street in the Pie Store. If the distance of Farco's position to the bottom of the Pie store is radical 62725 away, b) what is the angle of depression from the top of the building to the corner to the nearest hundreth?
Solutions
1.
2.
Tuesday, March 4, 2014
I/D #2: Unit O - How can we derive the patterns for our special right triangles?
INQUIRY ACTIVITY SUMMARY
1. 30-60-90 Triangle
The 30-60-90 triangle comes from an equilateral triangle, which has equilateral sides with equiangles of 60 degree. The equilateral triangle is cut in half, making: (2) triangles, (2) 30 degree angles, and (2) 90 degree angle. To explain the relationship between the sides and angles, an equilateral triangle with the sides of 1 can be used. If the process would be applied to this example, and focusing on ONE of the two triangle, the angles would remain the same as mentioned. The side cut in half will equal to 1/2, while the other untouched side (the hypotenuse in this case) would remain as 1. However, one side remains missing with its value, and in order to find it the Pythagorean Theorem (a^2 + b^2 = c^2) is used.
Using the Pythagorean Theorem, c would be the hypotenuse while a and b would be the other two sides that are across the 30 and 60 degree angles. Side a would equal to 1/2 while c would equal to 1. Plug those into the equation to solve for b will result with b equaling to radical 3 over 2. Across the 30 degree angle, a= 1/2; across the 60 degree angle, b= radical 3/2; and across the 90 degree angle c= 1. The relationship for a 30-60-90 triangle is 1/2-radical 3/2-1. HOWEVER, not all triangles are the same and not all derive from an equilateral triangle with the side of 1. The variable n can be used on the relation to represent any value for a triangle, creating (n/2)-n(radical 3)/2-(n). (And to make life simpler, the fraction can be removed my multiplying the relation by 2 to make (n)-n(radical 3)-2n..
2. 45-45-90 Triangle
The 45-45-90 triangle comes from cutting a square directly in half diagonally. Cutting the square diagonally in half will create: (2) triangles, (4) 45 degree angles, and (2) 90 degree angles. To explain the relationship between the sides and angles, a square with the sides of 1 can be used. Cutting the square and focusing on ONE triangle, the right triangle created will have two sides with the length of 1 with a missing hypotenuse. The sides of 1 will be across the 45 degree angles, and the hypotenuse can be found using the Pythagorean Theorem.
With the Pythagorean Theorem, side a will be 1 while side b will also be 1. After squaring the sides, adding them, and finding the square root, the hypotenuse will equal to radical 2. So a 45-45-90 triangle has a relationship of 1-1-radical 2. HOWEVER, not all triangles will derive from a square of 1. As in the 30-60-90 triangle, n can be used on the relation to represent any value of the triangle, creating (n)-(n)-n(radical 2).
INQUIRY ACTIVITY REFLECTION
“Something I never noticed before about special right triangles is…” that these special right triangles derived from equilateral triangles with equiangles. The 45-45-90 right triangle came from a square, a quadrilateral with the same length in sides and had (4) 90 right angles. The 30-60-90 right triangle derives from an equilateral triangle that had (3) 60 degree angles.
“Being able to derive these patterns myself aids in my learning because…” it gives me a better understanding where the numbers and relations actually come from and now know that a mathematician didn't randomly chose numbers. Also, in case I forget the relation during an important test, I could remember where and how these patterns are derived from.
1. 30-60-90 Triangle
2. 45-45-90 Triangle
With the Pythagorean Theorem, side a will be 1 while side b will also be 1. After squaring the sides, adding them, and finding the square root, the hypotenuse will equal to radical 2. So a 45-45-90 triangle has a relationship of 1-1-radical 2. HOWEVER, not all triangles will derive from a square of 1. As in the 30-60-90 triangle, n can be used on the relation to represent any value of the triangle, creating (n)-(n)-n(radical 2).
INQUIRY ACTIVITY REFLECTION
“Something I never noticed before about special right triangles is…” that these special right triangles derived from equilateral triangles with equiangles. The 45-45-90 right triangle came from a square, a quadrilateral with the same length in sides and had (4) 90 right angles. The 30-60-90 right triangle derives from an equilateral triangle that had (3) 60 degree angles.
“Being able to derive these patterns myself aids in my learning because…” it gives me a better understanding where the numbers and relations actually come from and now know that a mathematician didn't randomly chose numbers. Also, in case I forget the relation during an important test, I could remember where and how these patterns are derived from.
Saturday, February 22, 2014
I/D #1: Unit N Concept 7 - How do SRT and UC relate?
INQUIRY ACTIVITY SUMMARY
The quadrant in which the original triangles from the beginning are draw in the first one. Depending on which quadrant the SRT is found, the value will change by its sign (positive or not). If the triangle is found on Quadrant II, then x from the coordinate pair will be negative. In Quadrant III, both x and y will be negative. Finally in Quadrant IV, only y will be negative. In the three pictures provided above, the triangles in a different location from Quadrant I will change in its sign due new location. In Quadrant I, all trig functions are positive, while in Quadrant II only sin and csc are positive and the rest negative. Meanwhile in Quadrant III, tan and cot are the only positive. Finally in the Quadrant IV, only cos and sec are positive.
INQUIRY ACTIVITY REFLECTION
1.) “THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS..." that special right triangles are actually beneficial and do a bit more than just being a perfect shape.
2.) “THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE..." I can refer back to it if I ever forget how to fill out the Unit Circle for the coordinate pairs section.
3.) “SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS…” how they relate to one another. I would have not have guessed that the coordinates on the Unit Circle could also come from different shape.
1.) 30 degrees
There were several part that were needed to be labeled and solved for. The hypotenuse is equaled to 1, and due to the rules of the 30,60,90 right triangle: the side adjacent to 30 (which is x on the pic) has a value of x radical 3; the side opposite (which is y) has a value of x; and the hypotenuse (r) as 2x. To find an actual number value, trig functions (sin and cos) must be used. The sin of 30 degrees solves for the opposite side (y) and is found with y/r; substituting the variables will result with x/2x and simplified will equal to 1/2. The cos of 30 degrees solves for the adjacent side (x) and is found with x/r; substituting the variables will result with x radical 3/2x and simplified will equal to radical 3/2. If the triangle were to be placed in a Unit Circle, with the origin being located at the labeled angle measure, three coordinate pairs will be needed and two are found with the values solved. The origin is (0,0), while only going across the x axis is (radical 3/2,0), and going directly above from the previous coordinate is (radical 3/2,1/2).
2.) 45 degrees
The instructions are the same but differ in the number values. Due to the 45,45,90 right triangle rules, x is the side adjacent to the given degree with value of x; the y is opposite to the angle but similar to the adjacent side as it, too, has the value of x; and the hypotenuse, r, is equaled to 1 and x radical 2. The trig functions of sin and cos are also used with sin 45 degrees = y/r and cos 45 = x/r. Solving for the sin and cos is similar with this SRT since the same values are used to substitute, which is x/x radical 2 and simplified equaling to radical 2/2. If the triangle were to be placed in the Unit Circle, with the origin being located at the labeled angle measure, there would be three coordinate pairs. The origin would be (0,0), shifting to the right would be at (radical 2/2,0), and moving directly upward will get to (radical 2/2,radical 2/2)
3.) 60 degrees
The instructions and requirements are as well the same from the previous triangles, and it will be closely identical to the 30 degree triangle. Same as before: x is adjacent but with the value of x; y would be opposite to the angle with the value of x radical 3; and the hypotenuse, r, is 1 and 2x. Sin and cos will be used to find the length of the sides. The sin of 60 degrees (y/r) would be x radical 3/2x and simplified will equal to radical 3/2. The cos of 60 degrees (x/r) would be x/2x and simplified will equal to 1/2. If the triangle were to be placed in a Unit Circle, with the origin being located at the labeled angle measure, three coordinate pairs will be (0,0) at the origin, (1/2,) going across the x axis, and (1/2, radical 3/2,) shifting directly up.
4. This activity helps derive the unit circle by figuring out the certain point of the circle with the special right triangles. Figuring out the lengths of these triangles, if placed on the circle, will every time give the same coordinate pairs that are points on the Unit Circle if done correctly.
5.)
The quadrant in which the original triangles from the beginning are draw in the first one. Depending on which quadrant the SRT is found, the value will change by its sign (positive or not). If the triangle is found on Quadrant II, then x from the coordinate pair will be negative. In Quadrant III, both x and y will be negative. Finally in Quadrant IV, only y will be negative. In the three pictures provided above, the triangles in a different location from Quadrant I will change in its sign due new location. In Quadrant I, all trig functions are positive, while in Quadrant II only sin and csc are positive and the rest negative. Meanwhile in Quadrant III, tan and cot are the only positive. Finally in the Quadrant IV, only cos and sec are positive.
INQUIRY ACTIVITY REFLECTION
1.) “THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS..." that special right triangles are actually beneficial and do a bit more than just being a perfect shape.
2.) “THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE..." I can refer back to it if I ever forget how to fill out the Unit Circle for the coordinate pairs section.
3.) “SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS…” how they relate to one another. I would have not have guessed that the coordinates on the Unit Circle could also come from different shape.
Sunday, February 9, 2014
RWA #1: Unit M Concepts 4-6 - Conic Sections in real life (parabola)
1. Definition of Parabola
According to Mrs. Kirch, the definition of a parabola is a set of all points that are equidistant from a point (focus) and a line (directrix). The definition can best be explained/viewed from an image used in this page linked here.
2. Properties:
Algebraically:
According to Mrs. Kirch, the definition of a parabola is a set of all points that are equidistant from a point (focus) and a line (directrix). The definition can best be explained/viewed from an image used in this page linked here.
2. Properties:
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In standard form, parabolas are written as (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) (however in the picture to the left, four ways are shown and the only difference is simply because of the negative sign) . The variables h and k is the vertex (origin) of the parabola as h represents x and k to y. (Notice how on the image to the top left, h is paired with x and k with y). Also, p has to be on the side that is NOT being squared. (The image on the left represents p with an a, which is the same thing as explained in the video from above). The number of units of p represents how far is the focus point and the directrix line of the graph. The sign of p, too, (if it is positive or negative) determines if the graph will go up or down or left or right. Meanwhile, the variable that is squared can determine the direction of the parabola: horizontal or vertical.
The image on the right is an example of a problem solving to rewrite the equation in standard form. The side being squared was found by solving-the-square and the binomial is by itself. In the example, 4 represents p (or a), and notice how it is on the other side of the squared binomial and sitting alone, undistributed to the other binomial. The video from above can help explain a bit more on rewriting to standard form.
Graphically:
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The variable that is squared, as mentioned before, determines the direction of the parabola. As represented on the image to the right, if x is squared then the graph will go vertically; and if y is squared, then the graph will go vertically. The sign of p determines another part of the direction: if positive, the graph will move positive units either up or right; if negative, then the graph will move negative units either down or left. The axis of symmetry is always a line that spits the parabola in half from the vertex. The directrix is a line that is perpendicular to the axis of symmetry. As shown on the image to the left, the directrix is not within the parabola, but found outside and is found by p units away from the vertex. The focus point is within the graph and is found with p units away from the vertex. The distance away from the focus to the vertex can also help determine how wide or narrow the parabola would be. If the focus is far, then the parabola would be wide; and if the focus is close, then the parabola would be narrow.
3. RWA:
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| (visual.merriam-webster.com/astronomical) |
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| (parkerlab.bio.uci.edu) |
In the real world, parabolas can be found within the science, astronomical department. The images above show a radio telescope which, according to Merriam Webster Dictionary, is an "instrument used to capture, concentrate and analyze radio waves emanating from a celestial body or a region of the celestial sphere." Radio waves hit the parabolic reflector (the curved, bowl-like part of the telescope) which has a parabola shape. The waves automatically hits to the center of the reflector to a secondary reflector (considered the focus) that sends the waves down to the receiver. Anywhere where a wave would hit the parabolic reflector (a position on the parabola), it will always then aim to the secondary reflector (focus) since it was structured to have the same distance.
4. Work Cited:
- Conic Parabola Video - http://youtu.be/r-KmkpxVtGg
- Standard From Equations (pic) - https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcRf8C72WS3K4tZSQoWR_RF9hTYjCHXWeyeweyXqYD0PtWD3Gk2uWw
- Parabola Graphs (pic) - http://img.docstoccdn.com/thumb/orig/41803311.png
- Standard Form Example (pic) - https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcQKIXbJTTqeqSGF8weMMOnzNdoSH7DA-jc1CG6zvl7_tQWSdcov
- Parabola Parts w/ graph (pic)- https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQYmQdKDw1dEQKMsjpDpEbOr84_BF6-hOHW9Y_bP83bhlhapZIm
- Merriam Webster parabolic reflector (pic) - http://visual.merriam-webster.com/astronomy/astronomical-observation/radio-telescope.php
- Radio telescope (pic) - http://parkerlab.bio.uci.edu/nonscientific_adventures/otherplacesinCA.htm
Monday, January 13, 2014
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