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Thread: three sides of a triangle

  1. #1
    Nov 2006

    three sides of a triangle

    Let $\displaystyle a , b , c $ be the three sides of a triangle , and let

    $\displaystyle \alpha , \beta , \gamma $, be the angles opposite them. If $\displaystyle a^2+b^2=1989 c^2$,


    $\displaystyle \frac{\cot \gamma}{\cot \alpha+\cot \beta}$
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  2. #2
    Super Member wingless's Avatar
    Dec 2007
    Huh.. that was long..

    $\displaystyle a^2 + b^2 = 1989c^2$
    First thing I thought about this equation was the law of cosines. We'll use it later as expected.

    $\displaystyle \frac{\cot \gamma}{\cot \alpha+\cot \beta} = ?$
    Let's start of with reducing this.

    $\displaystyle \frac{\cot \gamma}{\cot \alpha+\cot \beta}$

    $\displaystyle \frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}}$

    $\displaystyle \frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\cos \alpha \sin \beta + \cos \beta \sin \alpha}{\sin \alpha \sin \beta}}$

    $\displaystyle \frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\sin (\alpha + \beta)}{\sin \alpha \sin \beta}}$

    As $\displaystyle \alpha + \beta + \gamma = \pi$,
    $\displaystyle \alpha + \beta = \pi - \gamma$

    $\displaystyle \frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\sin (\pi - \gamma)}{\sin \alpha \sin \beta}}$

    $\displaystyle \frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\sin \gamma}{\sin \alpha \sin \beta}}$

    $\displaystyle \frac{\cos \gamma \sin \alpha \sin \beta}{\sin \gamma \sin \gamma}$

    Law of sines tell us that $\displaystyle \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$

    From here we get $\displaystyle \frac{\sin \alpha}{\sin \gamma} = \frac{a}{c}$ and $\displaystyle \frac{\sin \beta}{\sin \gamma} = \frac{b}{c}$.

    By using them,
    $\displaystyle \frac{\cos \gamma \not\sin \not\alpha \not\sin \not\beta}{\not\sin \not\gamma \not\sin \not\gamma}\cdot \frac{a}{c} \cdot \frac{b}{c}$

    $\displaystyle \cos \gamma \cdot \frac{a.b}{c^2}$

    It's simple enough now. Time to find $\displaystyle \cos \gamma$ by the law of cosines..

    $\displaystyle a^2 + b^2 = 1989c^2$

    By the law,
    $\displaystyle c^2 = a^2 + b^2 - 2a.b \cos \gamma$

    $\displaystyle c^2 = 1989c^2 - 2a.b \cos \gamma$

    $\displaystyle 1988c^2 = 2a.b\cos\gamma$

    $\displaystyle 994c^2 = a.b\cos\gamma$

    $\displaystyle \cos\gamma = \frac{994c^2}{a.b}$

    We got $\displaystyle \cos\gamma$ now. By using it,

    $\displaystyle \cos \gamma \cdot \frac{a.b}{c^2} = ?$

    $\displaystyle \frac{994c^2}{a.b} \cdot \frac{a.b}{c^2}$

    $\displaystyle \boxed{994}$
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