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Math Help - Pyramid

  1. #1
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    Unhappy Pyramid

    Hello guys, me and my friends are desperate.. We can't solve this, any ideas? Thanks!
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  2. #2
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    Re: Pyramid

    Introduce two notations and write two equations for the tangents of the angles shown.
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  3. #3
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    Re: Pyramid

    Thanks for the reply
    We have tried to find x , h and then solve for h
    tan (40.3) = h / x + 200
    tan (40.27) = h / x + 100
    then we found x and solved for h but we got numbers like 0.525 which seem impossible to represent a high
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  4. #4
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    Re: Pyramid

    First, it's 46.27, not 40.27. Second, your equations are wrong because of the order of operations. Last, make sure that the tangent function interprets its arguments as given in degrees, not radians. The correct system gives the height between 440 and 470 feet.
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  5. #5
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    Re: Pyramid

    Sorry about the mistake, we have rearranged the equation to be: (x + 100)tan(46.27) = (x + 200)tan(40.3) we are trying to find the height
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  6. #6
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    Re: Pyramid

    Yes, this gives x(tan(46.27) - tan(40.3)) = 200 * tan(40.3) - 100 * tan(46.27), i.e.,

    x=\frac{200\tan40.3 - 100\tan46.27}{\tan46.27 - \tan40.3}\approx329.87

    Then h = (x + 100) * tan(46.27).
    Thanks from Imonars
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  7. #7
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    Re: Pyramid

    h = 449.3
    Thank you we got it!!
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  8. #8
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    Re: Pyramid

    Strictly speaking, if we round the height to one decimal digit, it should be 449.4 because the following digit is 6.
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  9. #9
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    Re: Pyramid

    Hello, Imonars!

    I would set it up like this:

    Code:
                                      o C
                                  * * |
                              *   *   |
                          *     *     |
                      *       *       |h
                  *         *         |
              * β         * α         |
        B o - - - - - - o - - - - - - o D
          :     100     A      x      :
    A man stands at A and sights the top of the pyramid C.
    The angle of elevation is \alpha = 46.27^o.

    He moves 100 feet away to point B\!:\;AB = 100.
    The angle of elevation is \beta = 40.3^o.

    The height of the pyramid is: h = CD.
    Let x = AD.


    In \Delta CDA\!:\;\tan\alpha \,=\, \frac{h}{x} \quad\Rightarrow\quad x \,=\, \frac{h}{\tan\alpha} .[1]

    In \Delta CDB\!:\;\tan\beta \,=\, \frac{h}{x+100} \quad\Rightarrow\quad x \,=\,\frac{h}{\tan\beta} - 100 .[2]


    Equate [2] and [1]: . \frac{h}{\tan\beta} - 100 \;=\;\frac{h}{\tan\alpha}

    Multipy by \tan\alpha\tan\beta\!:\;h\tan\alpha - 100\tan\alpha\tan\beta \:=\:h\tan\beta

    . . h\tan\alpha - h\tan\beta \:=\:100\tan\alpha\tan\beta

    . . h(\tan\alpha - \tan\beta) \:=\:100\tan\alpha\tan\beta

    . . . . . . . . . . . . h \:=\:\frac{100\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}


    Therefore: . h \;=\;\frac{100\tan46.27^o\tan40.3^o}{\tan46.27^o - \tan40.3^o} \;=\; 449.364461\hdots
    Thanks from Imonars
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  10. #10
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    Re: Pyramid

    That's how we got it, we handed in the paper 2 hours ago, the work is the same Thanks a alot
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