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Thread: Max/Min Problems with Trig.

  1. #1
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    Max/Min Problems with Trig.

    Hi

    Could someone please give me hints 4 the questions circled in the attachment? I have no clue how to start to arrive at those answers.

    Thanx a lot.
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  2. #2
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    Hello xwrathbringerx
    Quote Originally Posted by xwrathbringerx View Post
    Hi

    Could someone please give me hints 4 the questions circled in the attachment? I have no clue how to start to arrive at those answers.

    Thanx a lot.
    1(a) Let $\displaystyle QR = x$. Then $\displaystyle QP = x\cos\alpha$.

    So $\displaystyle x(1+\cos\alpha) = 2$ (= total length of wire)

    And $\displaystyle PR = x\sin\alpha$

    Then use $\displaystyle A = \tfrac12.PR.QP$


    2(a) Draw vertical lines through the top vertices of the trapezium to the base. Then using the right-angled triangles thus formed, the width of the channel across the top is $\displaystyle 4+ 4\cos\theta + 4\cos\theta = 4 + 8\cos\theta$, and the depth of the channel $\displaystyle = 4\sin\theta$.

    Then use the formula:

    Area of a trapezium =$\displaystyle \tfrac12$(sum of parallel sides) $\displaystyle \times$ (perpendicular distance between them)

    $\displaystyle =\tfrac12(4 + 4 + 8\cos\theta)\times 4\sin\theta$

    $\displaystyle = ...$


    Grandad
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  3. #3
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    Thumbs up

    Last edited by xwrathbringerx; Oct 2nd 2009 at 05:24 PM.
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