Fórum de Matemática
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In a triangle ABC, if BC =1 , sin (A/2) = x1 , sin (B/2)=x2 , cos (A/2) = y1 , cos(B/2)=y2 and (x1/x2)^2006 = (y1/y2)^2007 , then the length of AC is

Hello, good evening,

Well, here we go. This topic is a little old but I'll try not to leave it without help.

We have:
\(sin(A) = sin(2 \cdot A/2) = 2 \cdot sin(A/2) \cdot cos(A/2) = 2 \cdot x_1 \cdot y_1\).
\(sin(B) = sin(2 \cdot B/2) = 2 \cdot sin(B/2) \cdot cos(B/2) = 2 \cdot x_2 \cdot y_2\).

By the law of sines:
\(\frac{AC}{sin(B)} = \frac{BC}{sin(A)} \Rightarrow \frac{AC}{2 \cdot x_2 \cdot y_2 } = \frac{1}{2 \cdot x_1 \cdot y_1}\)

then:

\(AC = \left( \frac{x_2}{x_1} \right) \cdot \left( \frac{y_2}{y_1} \right)\).

Now you can use the provided relationships between \(x_1, x_2, y_1\) and \(y_2\) and try to finish the exercise. Ok?

_________________
Fraol
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