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Thread: Percentage error in Area of triangle

  1. #1
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    Percentage error in Area of triangle

    Hello, I'm stuck on the following question:

    Percentage error in Area of triangle-triangle-question-percentage-error.png

    I only get as far as taking natural logs and then differentiation it, but have no idea on how to carry on:

    ln(S) = ln(b)+ln(c)+ln(sinA)

    Assuming that b and c are equal to 1, I get:

    1/S dS = 1/b db + 1/c dc + cot A (sub in [A=pi/4])

    That's what I gather from the text book but then I hit dead end.

    Any idea on what I'm doing wrong or whether I'm completely wrong from the off?
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  2. #2
    MHF Contributor
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    Re: Percentage error in Area of triangle

    You are given that $\displaystyle \frac{1}{b}\ db$ and $\displaystyle \frac{1}{c}\ dc$ are both 5%. For the third term, you correctly differentiated

    $\displaystyle d(\log(\sin{A}))=(\cot{A})dA$

    but you are given $\displaystyle \frac{dA}{A}$, so you need:

    $\displaystyle d(\log(\sin{A}))=(\cot{A})A\ \frac{dA}{A}$.

    So $\displaystyle \frac{dA}{A}$ is 5%, A is $\displaystyle \frac{\pi}{4}$ and $\displaystyle \cot{A}$ is 1.

    - Hollywood
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