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Thread: How to find the altitude and size of an object atop a building

  1. #11
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    Quote Originally Posted by stapel View Post
    One can find the height of the top of the antenna but, no, nothing else can be determined.
    the height of the top of the antenna = 1.06233743 miles !!!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  2. #12
    Quote Originally Posted by stapel View Post
    One can find the height of the top of the antenna but, no, nothing else can be determined.
    Darn English, I should have said length of the antenna.
    Last edited by ecoo; 06-30-2016 at 12:17 PM.

  3. #13
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Subhotosh Khan View Post
    the height of the top of the antenna = 1.06233743 miles !!!!
    I'm getting something similar. For clarity, I'm using the following derived diagram:

    Code:
    diagram:
             *
            /|\
           / | \
         /  h|  \
        /    |   \
      /      |    \
     /60     |   70\ 
    *--------*------*
    |<--y--->|<---->|
              5280-y
    . . . . .[tex]\dfrac{h}{y}\, =\, \tan(60^{\circ})[/tex]

    . . . . .[tex]\dfrac{h}{5,280\, -\, y}\, =\, \tan(70^{\circ})[/tex]

    Solving for "y=":

    . . . . .[tex]\dfrac{h}{\tan(60^{\circ})}\, =\, y[/tex]

    . . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, =\, 5,280\, -\, y[/tex]

    . . . . .[tex]\dfrac{h}{\tan(70^{\circ})}\, -\,5,280\, =\, -y[/tex]

    . . . . .[tex]y\, =\, 5,280\, -\, \dfrac{h}{\tan(70^{\circ})}[/tex]

    This allows me to form an equation only in "h", which is the only variable I care about anyway:

    . . . . .[tex]5,280\, -\, \dfrac{h}{\tan(70^{\circ})}\, =\, \dfrac{h}{\tan(60^{\circ})}[/tex]

    . . . . .[tex]5,280\, =\, h\, \left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)[/tex]

    . . . . .[tex]\dfrac{5,280}{\left(\dfrac{1}{\tan(60^{\circ})}\, +\, \dfrac{1}{\tan(70^{\circ})}\right)}\,=\, h[/tex]

    So the result is an utterly unreasonable value.

  4. #14
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    Using Law of Sines once, then the 30-60-90 triangle,
    directly makes it 1.062337425275....as per Subhotosh...

    k = SIN(70) / SIN(50)

    h = k/2 * SQRT(3)
    Last edited by Denis; 06-30-2016 at 09:22 PM.
    I'm just an imagination of your figment !

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