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Math Help - Problem Solving

  1. #1
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    Problem Solving

    An observer is on the bank of a river at A, opposite a tower TF. His task is simply to find the height of the tower. He cannot get to the tower because of the river and so he has to do this from where he is. He is equipped only with an inclinometer, a marking pole and a measuring tape.

    How can he measre the height of the tower? Can he still do it if he has only the marking pole and the measuring tape?
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  2. #2
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    Hello, matgrl!

    This is a classic problem in Trigonometry.


    An observer is on the bank of a river at A, opposite a tower TF.
    His task is simply to find the height of the tower.
    He cannot get to the tower because of the river,
    so he has to do this from where he is.
    He is equipped only with an inclinometer, a marking pole and a measuring tape.

    How can he measure the height of the tower?
    Can he still do it with only the marking pole and the measuring tape? . I don't think so.
    Code:
                              * T
                            */|
                          * / |
                        *  /  |
                      *   /   |
                    *    /    | h
                  *     /     |
                *      /      |
              *       /       |
            * β      / α      |
        B * - - - - * - - - - * F
              50    A    x

    The tower is h = TF.
    The observer is at \,A. .Let x = AT.
    He finds that the angle of elevation to the top of the tower is \,\alpha.

    He walks a distance directly away from the tower, say, 50 feet, to point \,B.
    At \,B, the angle of elevation to the top of the tower is \,\beta.


    In right triangle TFB\!:\;\;\tan\beta \:=\:\dfrac{h}{x+50} \quad\Rightarrow\quad x \:=\:\dfrac{h-50\tan\beta}{\tan\beta} .[1]

    In right triangle TFA\!:\;\;\tan\alpha \:=\:\dfrac{h}{x} \quad\Rightarrow\quad x \:=\:\dfrac{h}{\tan\alpha} .[2]


    Equate [1] and [2]: . \dfrac{h-50\tan\beta}{\tan\beta} \:=\:\dfrac{h}{\tan\alpha}

    . . . . . . . . . \tan\alpha(h-50\tan\beta) \;=\;h\tan \beta

    . . . . . . h\tan\alpha - 50\tan\alpha\tan\beta \;=\;h\tan\beta

    . . . . . . . . . . . h\tan\alpha - h\tan\beta \;=\;50\tan\alpha\tan\beta

    . . . . . . . . . . . h(\tan\alpha - \tan\beta) \;=\;50\tan\alpha\tan\beta

    . . . . . . . . . . . . . . . . . . . . . h \;=\;\dfrac{50\tan\alpha\tan\beta}{\tan\alpha - \tan\beta}
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