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# Let $ABC$ be a triangle such that $\angle ACB=\large\frac{\pi}{6}$.Let $a,b$ and $c$ denote the lengths of the sides opposite to $A,B$ and $C$ respectively.The value of $x$ for which $a=x^2+x+1,b=x^2-1$ and $c=2x+1$ is

$\begin{array}{1 1}(a)\;-(2+\sqrt 3)&(b)\;1+\sqrt 3\\(c)\;2+\sqrt 3&(d)\;4\sqrt 3\end{array}$

Can you answer this question?

$\large\frac{\sqrt 3}{2}=\frac{(x^2+x+1)^2+(x^2-1)^2-(2x+1)^2}{2(x^2+x+1)(x^2-1)}$
$\Rightarrow (x+2)(x+1)(x-1)x+(x^2-1)^2=\sqrt 3(x^2+x+1)(x^2-1)$
$\Rightarrow x^2+2x+(x^2-1)=\sqrt 3(x^2+x+1)$
$\Rightarrow (2-\sqrt 3)x^2+(2-\sqrt 3)x-(\sqrt 3+1)=0$
$x=(1+\sqrt 3)$
Hence (b) is the correct option.
answered Nov 18, 2013
edited Mar 21, 2014