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Tutor profile: Mary T.

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Mary T.
Teacher of Mathematics
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Questions

Subject: Trigonometry

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Question:

The study of triangle relationships is useful in determining unknown distances. Right triangles, in particular, can provide answers to unknowns with only two pieces of information. Suppose you need to replace a guy wire (a tension cable for stability) which is attached to a utility pole. The guy wire is fastened to the ground 27 meters from the base of the utility pole. The angle of elevation of the wire is 50 degrees. What is the length of the guy wire?

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Mary T.
Answer:

The utility pole, guy wire, and ground form a right triangle. The length of the guy wire is the unknown hypotenuse of the triangle. In relation to the angle of elevation, the ground is the adjacent side. The trigonometric function that relates this angle, adjacent side, and hypotenuse is cosine: cos (50) = 27 / hyp. Cross products show that hyp = 27 / cos (50). Using a calculator set in degree mode, hyp = 42. The length of the guy wire is approximately 42 meters.

Subject: Geometry

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Question:

In city planning, the points of concurrency of a triangle may assist in determining locations of specific buildings. For instance, only one restroom facility may be build in a city park. The location must be accessible to the 3 major locations of the park: a baseball field, basketball court, and children's playground. How does the city determine where the restroom facilities should be located so that it is equidistant to the three major locations?

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Mary T.
Answer:

The baseball field, basketball court, and children's playground represent the vertices of a triangle. In order to find the ideal location of the restroom facilities, we would need to determine the circumcenter of the triangle formed by the three locations. The circumcenter is found by generating the perpendicular bisectors of each side of the triangle. This means we must measure the sides of the triangle formed by the three locations in order to find the midpoint of each side. From there, a line must be constructed that passes through the midpoint of each side and is also perpendicular to that side. Once completed, the perpendicular bisectors of each side of the triangle formed by the three locations will intersect at the perfect location for the restroom facilities and be equidistant from each location.

Subject: Algebra

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Question:

A common topic in Algebra is Systems of Equations. They are useful in situations that involve two variables. Suppose you were told the following: Farmer Joe looked out at his field where his cows and chickens were grazing. He counted 20 heads and 70 legs. How many cows and how many chickens does Farmer Joe have in the field?

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Mary T.
Answer:

The two variables are cows (x) and chickens (y). The relationship between the two is the total number of heads and total number of legs. Additionally, we know that cows have 4 legs whereas chickens have 2. A system of equations to represent this situation is x + y = 20 (the total number of cows and chickens is 20, as revealed by the head count) and 4x + 2y = 70 (the total number of legs is 70; cows have 4, chickens have 2.) There are 4 different ways to solve this system. 1) elimination: the equations are set up in standard form, with x and y on one side of the equation. If we can eliminate a variable by forming a zero pair then the equation can be solved. In this case, the top equation would be multiplied by -2, creating -2x - 2y = -40. Adding the equations to each other forms a zero pair with the y variables since 2+ -2 is 0. We are left with 2x = 30. Solving this one step equation by diving yields x = 15. There are 15 cows. Since there are 20 animals in total, 5 must be chickens. 2) substitution: one equation must be solved for one variable. Neither is set up this way, be we can easily solve the first equation for x or y. Solve for x yields x = 20 - y. Now we can substitute this equation into the second equation where we see x. 4(20 - y) + 2y = 70. Solving this equation by distributing and further methods of equation solving yields y = 5. There are 5 chickens. Since there are 20 animals in total, 15 must be cows. 3) graphing: both equations must be solved for y in order to graph. The first equation becomes y = 20 - x and the second is y = 35 - 2x. After graphing, the intersection of the two lines is the solution to the system. The intersection point would be (15, 5) meaning there are 15 cows (x) and 5 chickens (y). 4) diagram: begin by drawing 20 circles representing the 20 heads. Both types of animals have at least 2 legs. You can draw two sticks on the circles to represent this. However, 20 circles with 2 sticks each means we only have 40 legs, yet we know there must be 70. Add on pairs of legs until the sum of legs reaches 70. At this point, count the circles with four sticks and that is the number of cows. Count the circles with two legs and that is the number of chickens.

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