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Math Help - Relation between cosines and lenghts of triangle

  1. #1
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    Relation between cosines and lenghts of triangle

    Prove that in any triangle we have the following relation \frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2}  }\cos{\frac{C}{2}}
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    Re: Relation between cosines and lenghts of triangle

    What's R? ...

    It can be useful to know that in a triangle:
    A+B+C=180
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    Re: Relation between cosines and lenghts of triangle

    R is radius of the circumscribed circle.
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    Re: Relation between cosines and lenghts of triangle

    RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }

    = 8s(area of triangle)/abc

    now using s = (a+b+c)/2 and 4(area of triangle)/abc = 1/R you get required result
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    Re: Relation between cosines and lenghts of triangle

    could you be more speciffic?
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    Re: Relation between cosines and lenghts of triangle

    about what, if you point out the problem
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    Re: Relation between cosines and lenghts of triangle

    Quote Originally Posted by waqarhaider View Post
    RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }
    about this
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    Re: Relation between cosines and lenghts of triangle

    cos(A/2) = sqrt[ s(s-a)/bc] , cos(B/2) = sqrt[ s(s-b)/ac] and cos(C/2) = sqrt[s(s-c)/ab]
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