1(x1x2)2007=(x3x4)2006
=>(sinA2/sinB2)2007=(x3x4)2006
Applying sine rule to L.H.S.
( kakb)2007=(x3x4)2006
=>( ab)2007=(x3x4)2006
=>( BCAC)2007=(x3x4)2006
Given that BC=1
( 1AC)2007=(x3x4)2006
=>AC=(x4x3)20062007
Not a dbt.
In triangle ABC, BC is of unit length.
Given sin\frac{A}{2}=x_1, sin\frac{B}{2}=x_2, cos \frac{B}{2}=x_4 & cos \frac{A}{2}=x_3.
Also Given that \left(\frac{x_1}{x_2} \right)^{2007}=\left(\frac{x_3}{x_4} \right)^{2006}.
Find the length of AC.
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