three sides of a triangle

**perash** · March 31st 2008, 09:46 AM

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

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

find

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

**wingless** · April 1st 2008, 07:22 AM

Huh.. that was long..

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{994}$

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