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Math Help - Two angles and the angle in between in 180 degrees; maximization

  1. #1
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    Thumbs up Two angles and the angle in between in 180 degrees; maximization

    Hi Forum!
    I have a question



    We are to find the x coordinate of P to maximize abp
    a(1,2) [use your imagination here, the graph is not right hahaha]
    b(-1,1)
    c(-1,0)
    d(1,0)

    Now, if we consider t as the x coordinate of P, we have:
    Tan(A)=2/1-t
    Tan(B)=1/1+t

    If we have to find Tan(C), how should we do it?
    A+B+C=180
    C=180-A-B
    Where do we go from there?

    We know that tan(180) is 0, but how are we going to use it?


    Thanks.
    Last edited by Zellator; June 14th 2011 at 06:35 AM. Reason: Correction for better understanding
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  2. #2
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    Opalg's Avatar
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    Re: Two angles and the angle in between in 180 degrees; maximization

    Quote Originally Posted by Zellator View Post
    Hi Forum!
    I have a question



    We are to find the x coordinate of P to maximize abp
    a(1,2) [use your imagination here, the graph is not right hahaha]
    b(-1,1)
    c(-1,0)
    d(1,0)

    Now, if we consider t as the x coordinate of P, we have:
    Tan(A)=2/1-t
    Tan(B)=1/1+t

    If we have to find Tan(C), how should we do it?
    A+B+C=180
    C=180-A-B
    Where do we go from there?

    We know that tan(180) is 0, but how are we going to use it?


    Thanks.
    \tan(180^\circ-C) = -\tan C, and so \tan C = -\tan(A+B) = -\frac{\tan A + \tan B}{1-\tan A\tan B}. Now take it from there ... .
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  3. #3
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    Re: Two angles and the angle in between in 180 degrees; maximization

    Quote Originally Posted by Opalg View Post
    \tan(180^\circ-C) = -\tan C, and so \tan C = -\tan(A+B) = -\frac{\tan A + \tan B}{1-\tan A\tan B}. Now take it from there ... .
    Hey Opalg!
    I remember seeing this formula somewhere, but I didn't know how to put in motion.
    So, it was easier than I thought. Thanks for your explanation!!!
    All the best!
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