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Math Help - Percent Error (Simple Problem?)

  1. #1
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    Oct 2008
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    Question Percent Error (Simple Problem?)

    Hi, I'm new here, but I was looking for some help with a Calc problem and here I am. Odd thing about this problem is we worked it out in class, but when checking the answer it seems wrong. I talked to my calc teacher and she's stumped too, so can anyone see what's wrong here?

    A surveyor is standing 50 feet from the base of a tree. He measures his angle of elevation to the top of the tree is 71.5 degrees. How accurately must he measure his angle of elevation to measure the within a 6% error.

    This is what I did. Set up a triangle and get

    h = 50tan(a)


    Differentiating the equation I wind up with


    dh/da = 50sec^2(a)


    Because we want dh/h to be less than 6% I set dh = .06h
    A little algebra gives me :

    da = dh/(50sec^2(a)) = .06h/(50*sec^2(a))
    <br />
da = (.06*50tan(a))/(50sec^2(a))<br />


    Cancelling the 50's and converting the

    sec^2 to 1/cos^2

    I get

    <br />
da = .06tan(a)cos^2(a)<br />

    Exact height at 71.5 degrees would come out to be 149.4342.

    Plugging in 71.5 my da comes out to be roughly .0181
    At first I thought this was degrees which gives no where near a 6% differences in the height. Then I thought it was radians. Converting .0181 from radians to degrees gives 1.03 degree difference. Applying this gives a 6.3% difference in height.. which is close, but it's still over my tolerance.

    I know it's rubbish but i did discover that 1 - .0181 = .9819 which added to my 71.5 degrees and applied to the height equation gives me the 6% exactly, but I can't think of any logical reason I would want to subtract my result from 1.

    Any help on this (admittedly long winded) post would be greatly appreciated. Thanks.
    Last edited by jackbunny; October 22nd 2008 at 12:09 PM. Reason: Poor spelling
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  2. #2
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    I hate to do it, but I'm going to give this a little bump in hopes someone will be able to help. Thanks.
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