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Math Help - Applied cosine rule problem.

  1. #1
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    Question Applied cosine rule problem.

    Q: The points A,B & C lie in a straight line with AB=x & BC=2x.
    Avertical tower OH is of height h and its base Olies in the same
    horizontal plane as ABC but not on the line ABC.
    The angle of elevation of H from A,B & C are alpha, beta & alpha
    respectively.

    (i) prove thah h^2(cot^2(alpha) - cot^2(beta)) = 2x^2
    ------------------------------------------------------------------
    So:
    (i) I start by taking the triangle AOB where AB = x and AHO & BHO
    (ii)Because I'm required to prove using cot I'm going to use tan.
    Therefore:
    tan (alpha) = h/AO
    AO = h/tan (alpha)
    AO = h*cot (alpha)
    tan (beta) = h/BO
    BO = h/tan (beta)
    BO = h*cot (beta)
    tan (alpha) = h/CO
    CO = h/tan (alpha)
    CO = h*cot (alpha)
    working out that in triangle CBO the angle is unkown called theta and triangle ABO the angle must be 180 - theta
    (iii) using cosine law with Cos (theta as the subject)
    CBO:
    Cos (theta) = {(2x)^2 + (BO)^2 - (CO)^2}/ 2*(2x)*(BO)
    ABO:
    Cos (180 -(theta)) = {(x)^2 + (BO)^2 - (AO)^2}/ 2*(x)*(BO)
    Therefore, Cos (180 -(theta)) = -cos (theta)
    therefore {(O) - (BO)^2 - (x)^2}/ 2*(x)*(BO)
    Replacing AO,BO & CO for the cot values these two equations shouldbe equal to one another. I don't seem to get the right answer after manipulating them though

    Can you show me how this is proven? There are other parts to the
    question but I think once I understand this I will be okay on my own.
    many thanks
    Last edited by dojo; February 7th 2010 at 05:15 AM.
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  2. #2
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    Hello, dojo!

    Sorry, I don't understand the problem.
    Some information is missing . . .


    The points A,B,C lie in a straight line with: . AB=x,\;BC=2x
    . .
    Is line AC horizontal?

    A vertical tower OH is of height h and its base O
    lies in the same horizontal plane as ABC, but not on the line ABC. .
    Really?

    The angle of elevation of H from A,B,C are \alpha,\;\beta,\;\alpha, resp.

    (i) Prove that: . h^2\left(\cot^2\alpha - \cot^2\beta\right) .
    . . . equals WHAT?

    Assuming AC is horizontal, this is all I get:


    Code:
                          H
                          o
                        **| *
                      * * |   *
                    *  *  |h    *
                  *   *   |       *
                *    *    |         *
              *     *     *O          *
            * α    * β               α  *
          o - - - o - - - - - - - - - - - o 
          A   x   B          2x           C
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  3. #3
    Member
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    Ah, yes good point!! the proof is to equal 2x^2 - I have added this to the question now. ABC is a prependicular to O, so in the same plane. Your diagram is correct.

    Do my workings seem correct?

    D
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