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Trigonometric Ratios Using PowerPoint

R
richrollo

The document is a student's presentation summarizing the steps to solve a trigonometry problem about determining the height of lights on a building from an observer's perspective. The presentation shows setting up the tangent ratio equation using the given information of the observer being 2 miles from the building at a 5 degree angle of elevation. The student uses trigonometric functions to isolate and solve for the variable, determining the height of the lights is approximately 900 feet above the ground.

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Trigonometric Ratios by Amy H. This PowerPoint presentation was made by one of my 7 th grade honors students. The assignment was to demonstrate the steps in solving a trigonometry problem of her choice.
The Question Suppose you are standing 2 miles away from a tall building and you see the lights on top of the building. The angle of elevation from you to the lights is 5 . To the nearest 100 feet, how far above the ground are the lights?
Picture the Problem 2 miles 5 X
Choose a Trig Function SINE of 5 COSINE of 5 TANGENT of 5 To calculate the height of the lights at the top of the building.
The Choice The only Trigonometric ratio that will work with the given information is the TAN of 5 . The tangent is the choice when the hypotenuse measure is missing. 2 miles 5 ?? X
Set-up of the Equation TAN of 5 = X 2 miles First, convert the 2 miles into feet (2 X 5,280) because the answer is needed to be to the nearest 100 feet. Now the equation becomes TAN of 5 = X 10,560 ft.
Function Translation Convert the TAN of 5 into a decimal using a calculator or a function chart. I choose to use Mr. Rollos function chart for my conversion. TAN of 5 = X 10,560 ft. Now becomes .08749 = X 10,560 ft.
Isolate the variable Multiply both sides of the equation by 10,560. .08749 = X 10,560 ft. (10,560) (10,560) Now becomes 923.8944 = X
Solution 923.8944 = X The question requested that the answer be rounded to the nearest 100 feet. Therefore 923.8944 = 900 feet
Conclusion The building lights are about 900 feet above the ground. 2 miles 5 ?? 900

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Trigonometric Ratios Using PowerPoint

  • 1. Trigonometric Ratios by Amy H. This PowerPoint presentation was made by one of my 7 th grade honors students. The assignment was to demonstrate the steps in solving a trigonometry problem of her choice.
  • 2. The Question Suppose you are standing 2 miles away from a tall building and you see the lights on top of the building. The angle of elevation from you to the lights is 5 . To the nearest 100 feet, how far above the ground are the lights?
  • 3. Picture the Problem 2 miles 5 X
  • 4. Choose a Trig Function SINE of 5 COSINE of 5 TANGENT of 5 To calculate the height of the lights at the top of the building.
  • 5. The Choice The only Trigonometric ratio that will work with the given information is the TAN of 5 . The tangent is the choice when the hypotenuse measure is missing. 2 miles 5 ?? X
  • 6. Set-up of the Equation TAN of 5 = X 2 miles First, convert the 2 miles into feet (2 X 5,280) because the answer is needed to be to the nearest 100 feet. Now the equation becomes TAN of 5 = X 10,560 ft.
  • 7. Function Translation Convert the TAN of 5 into a decimal using a calculator or a function chart. I choose to use Mr. Rollos function chart for my conversion. TAN of 5 = X 10,560 ft. Now becomes .08749 = X 10,560 ft.
  • 8. Isolate the variable Multiply both sides of the equation by 10,560. .08749 = X 10,560 ft. (10,560) (10,560) Now becomes 923.8944 = X
  • 9. Solution 923.8944 = X The question requested that the answer be rounded to the nearest 100 feet. Therefore 923.8944 = 900 feet
  • 10. Conclusion The building lights are about 900 feet above the ground. 2 miles 5 ?? 900