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Math Help - Prove for trianlge ABC

  1. #1
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    Prove for trianlge ABC

    1. \cot C=\frac{a}{c}\csc B-\cot B
    I have not idea where to start here.

    2. abc=4\bigtriangleup R where R is the radius of the circumcircle, and \bigtriangleup is the area of the triangle.
    I used \bigtriangleup=\sqrt{s(s-a)(s-b)(s-c)} where s=\frac{1}{2}(a+b+c) and got the value for \bigtriangleup=\frac{1}{4}\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2} which doesn't have a square root. Now I'm stuck
    Thanks
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  2. #2
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    Hello arze
    Quote Originally Posted by arze View Post
    1. \cot C=\frac{a}{c}\csc B-\cot B
    I have not idea where to start here.

    2. abc=4\bigtriangleup R where R is the radius of the circumcircle, and \bigtriangleup is the area of the triangle.
    I used \bigtriangleup=\sqrt{s(s-a)(s-b)(s-c)} where s=\frac{1}{2}(a+b+c) and got the value for \bigtriangleup=\frac{1}{4}\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2} which doesn't have a square root. Now I'm stuck
    Thanks
    1. Use the Sine Rule on the RHS, by writing
    \frac{a}{c}=\frac{\sin A}{\sin C}
    and write \csc B and \cot B in terms of \sin B and \cos B.

    So:
    \frac{a}{c}\csc B-\cot B =\frac{\sin A}{\sin C\sin B}-\frac{\cos B}{\sin B}
    =\frac{\sin A-\sin C\cos B}{\sin C\sin B}
    Now use the fact that \sin A = \sin (180^o-A) = \sin(B+C). Expand this, and then simplify the numerator. Can you finish this now?

    2. Use \Delta = \tfrac12bc\sin A, and \frac{a}{\sin A}=2R and you're done.

    Grandad
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