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Math Help - Find Height of Helicopter

  1. #1
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    Find Height of Helicopter

    Two observers measure, at the same time, the angle of elevation of a helicopter. One angle is measured as 25 degrees, the other angle is measured as 40 degrees. If the observers are 100 feet apart from each other and the helicopter lies over the line joining them, how high is the helicopter?
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  2. #2
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    Quote Originally Posted by magentarita View Post
    Two observers measure, at the same time, the angle of elevation of a helicopter. One angle is measured as 25 degrees, the other angle is measured as 40 degrees. If the observers are 100 feet apart from each other and the helicopter lies over the line joining them, how high is the helicopter?
    A large well labeled diagram is essential. At the risk of a 12 pt purple coloured attack, I must ask:

    Have you drawn one?

    Let the observers be at A and B. Let the helicopter be at C. Let the perpendicular line from C meet AB at D.

    Let AD = x so that DB = 100 - x. Let DC = h = height of helicopter.

    From right-triangle ADC: \tan 25^0 = \frac{h}{x} \Rightarrow x \tan 25^0 = h .... (1)

    From right-triangle BDC: \tan 40^0 = \frac{h}{100-x} \Rightarrow (100 - x) \tan 40^0 = h .... (2)

    Equate equations (1) and (2) and solve for x (your value for x should be more accurate than the accuracy required for the value of h). Substitute the value of x into equation (1) and solve for h.
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  3. #3
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    You can also use Sine Law and then use either one of the triangles to solve for the altitude.

    At the risk of a 12 pt purple coloured attack
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  4. #4
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    Smile Great set up...

    Quote Originally Posted by mr fantastic View Post
    A large well labeled diagram is essential. At the risk of a 12 pt purple coloured attack, I must ask:

    Have you drawn one?

    Let the observers be at A and B. Let the helicopter be at C. Let the perpendicular line from C meet AB at D.

    Let AD = x so that DB = 100 - x. Let DC = h = height of helicopter.

    From right-triangle ADC: \tan 25^0 = \frac{h}{x} \Rightarrow x \tan 25^0 = h .... (1)

    From right-triangle BDC: \tan 40^0 = \frac{h}{100-x} \Rightarrow (100 - x) \tan 40^0 = h .... (2)

    Equate equations (1) and (2) and solve for x (your value for x should be more accurate than the accuracy required for the value of h). Substitute the value of x into equation (1) and solve for h.
    Great set up here.
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