Question

Sean

a,

b,

c

los tres lados de un triángulo, y sean

\alpha,

\beta,

\gamma

los ángulos opuestos a ellos. Si

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

halla

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

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=1989c^2,

find

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

Respuesta

Simplifica

\cot\alpha+\cot\beta

usando

\alpha+\beta=\pi-\gamma

Simplify

\cot\alpha+\cot\beta

using

\alpha+\beta=\pi-\gamma

Usa la ley de los senos para los factores de seno resultantes y la ley de los cosenos para

\cos\gamma

Use the Law of Sines for the resulting sine factors and the Law of Cosines for

\cos\gamma

Primero,

\begin{aligned}\cot\alpha+\cot\beta&=\frac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta}\\&=\frac{\sin\gamma}{\sin\alpha\sin\beta}.\end{aligned}

Por consiguiente, la razón pedida es

\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=\frac{ab\cos\gamma}{c^2},

donde se usó la ley de los senos en la última igualdad. Por la ley de los cosenos,

\begin{aligned}2ab\cos\gamma&=a^2+b^2-c^2\\&=1988c^2.\end{aligned}

Por lo tanto, la razón es

\frac{1988}{2}=994.

First,

\begin{aligned}\cot\alpha+\cot\beta&=\frac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta}\\&=\frac{\sin\gamma}{\sin\alpha\sin\beta}.\end{aligned}

Hence the desired ratio is

\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=\frac{ab\cos\gamma}{c^2},

where the Law of Sines was used in the final equality. By the Law of Cosines,

\begin{aligned}2ab\cos\gamma&=a^2+b^2-c^2\\&=1988c^2.\end{aligned}

Therefore the ratio is

\frac{1988}{2}=994.