SOLUTION: You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a point directly under the goal post is 72° and the b

Algebra ->  Trigonometry-basics -> SOLUTION: You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a point directly under the goal post is 72° and the b      Log On


   



Question 1042621: You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a point directly under the goal post is 72° and the blimp will be directly above the 50 yard line.
a. Which trigonometric ratio would you use to calculate how high the blimp will be above the 50 yard line?
b. How high above the ground is the blimp?
c. In order to be able to read the advertisement on the side of the blimp the highest the blimp can be is 150 feet.
Will the fans be able to read the advertisement? If not, what possible angle of elevation could we use?
d. What is the exact angle if the blimp is at 150 feet?
Please answer thoroughly. Thank you for your assistance.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a point directly under the goal post is 72° and the blimp will be directly above the 50 yard line.
Sketch a right triangle with base = 50 yd, base angle = 72 deg, height = h
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a. Which trigonometric ratio would you use to calculate how high the blimp will be above the 50 yard line?
sin(72 deg) = h/50
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b. How high above the ground is the blimp?
h = 50*sin(72) = 47.55 yd
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c. In order to be able to read the advertisement on the side of the blimp the highest the blimp can be is 150 feet.
47.55 yd = 47.55*3 ft = 142.66 ft
Will the fans be able to read the advertisement?:: Yes
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d. What is the exact angle if the blimp is at 150 feet?
tan^-1(150/50) = tan^-1(3) = 71.56 degrees
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Cheers,
Stan H.
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