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Math Help - 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.
    Attached Thumbnails Attached Thumbnails Max/Min Problems with Trig.-001.jpg  
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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 QR = x. Then QP = x\cos\alpha.

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

    And PR = x\sin\alpha

    Then use 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 4+ 4\cos\theta + 4\cos\theta = 4 + 8\cos\theta, and the depth of the channel = 4\sin\theta.

    Then use the formula:

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

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

    = ...


    Grandad
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  3. #3
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    Thumbs up

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