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Math Help - Real World Problem

  1. #1
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    Post Real World Problem

    Hello,

    I work for a small manufacturing firm. We are trying to work out the technical details of how to figure out whether a piece of hardware for a military gun is right or not. This has led me down the road of some algebra that has me stumped.

    What I have is:
    K/tan(B) = (0.108-K)/tan(A)

    All I need to do is to manipulate this down to a "K=" formula, then I can program this into our measurement equipment and away we'll go. But right now I am stuck.

    I will be grateful for any help that might be available.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by mcinteer View Post
    Hello,

    I work for a small manufacturing firm. We are trying to work out the technical details of how to figure out whether a piece of hardware for a military gun is right or not. This has led me down the road of some algebra that has me stumped.

    What I have is:
    K/tan(B) = (0.108-K)/tan(A)

    All I need to do is to manipulate this down to a "K=" formula, then I can program this into our measurement equipment and away we'll go. But right now I am stuck.

    I will be grateful for any help that might be available.
    The basic rule you need is \frac{a+ b}{c}= \frac{a}{c}+ \frac{b}{c}.

    \frac{K}{tan(B)}= \frac{0.108- K}{tan(A)}= \frac{0.108}{tan(A)+ \frac{K}{tan(A)}}
    Subtract \frac{K}{tan(A)} from both sides:
    \frac{K}{tan(B)}- \frac{K}{tan(A)}= K\left(\frac{1}{tan(B)}-\frac{1}{tan(A)}\right)= \frac{0.108}{tan(A)}
    Get common denominators to subtract those fractions:
    \frac{tan(A)}{tan(A)tan(B)- \frac{tan(B)}{tan(A)tan(B)}= \frac{tan(A)- tan(B)}{tan(A)tan(B)}
    so
    K\frac{tan(A)- tan(B)}{tan(A)tan(B)}= \frac{0.108}{tan(A)}
    and divide both sides by that fraction on the left:
    K= \frac{0.108}{tan(A)}\frac{tan(A)tan(B)}{tan(A)- tan(B)}
    K= \frac{0.108(tan(A)- tan(B)}{tan(B)}
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