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Math Help - Law of Tangents

  1. #1
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    Law of Tangents

    Hey MHF, today in class we took Melloweid's formula, then the teacher gave us this equation and only two students were able to solve it
    I'm 100% sure I'm going to have to use Melloweid's formula to derive it.

    Law of Tangents For any triangle, derive the Law of Tangents.


    Any ideas?
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  2. #2
    Junior Member Barioth's Avatar
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    Re: Law of Tangents

    Hi the Melloweird's formula are given there

    Mollweide's formula - Wikipedia, the free encyclopedia

    if you take the second one and divide it by the firt one you end up with

    \frac{(a-b)}{(a+b} = \tan{(\frac{1}{2}(A-B)}\tan{(\frac{\theta}{2})}
    we know that
     A+B = 180 - \theta
    so
     \tan{(\frac{\theta}{2})} = \tan{(90 -\frac{A+B}{2})}

    The tricky part is the find out Why in the world
     \tan{(90 -\frac{A+B}{2})} = \arctan{(\frac{A+B}{2}})

    But this is the funny part so I leave it to you!
    Last edited by Barioth; January 16th 2013 at 04:07 PM.
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  3. #3
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    Re: Law of Tangents

    Lol, I will give it a thought and see how it goes, if I find it I will post the results. If not, then know that I have failed
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  4. #4
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    Re: Law of Tangents

    I'm gonna call it a night
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  5. #5
    Junior Member Barioth's Avatar
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    Re: Law of Tangents

    dang I think I made a mistake there! Hold on i'll change few thing..

    there it should be fine. Sorry about that.
    Last edited by Barioth; January 16th 2013 at 04:14 PM.
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  6. #6
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    Re: Law of Tangents

    Haha no way, I couldn't.. I give up :/
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  7. #7
    Junior Member Barioth's Avatar
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    Re: Law of Tangents

    Here is how you solve it

    sin(a-b) = sin a*cos b - sin b*cos a

    Let a = pi/2 and b = x = (A+B)/2

    sin ( /2 - x) = sin /2*cos x - sin x* cos /2

    since sin pi/2 = 1 and cos pi/2 = 0

    sin ( pi/2 - x) = 1*cos x - sin x*0

    sin ( pi/2 - x) = cos x

    note: I personaly just write down my trigo circle and see it eazier from there.

    Have a good day
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