# three sides of a triangle

• Mar 31st 2008, 09:46 AM
perash
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 $\displaystyle a^2+b^2=1989 c^2$,

find

$\displaystyle \frac{\cot \gamma}{\cot \alpha+\cot \beta}$
• Apr 1st 2008, 07:22 AM
wingless
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,

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

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

$\displaystyle \boxed{994}$