Let ABC be a triangle such that angle ACB = f | vedclass

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(B) Using the Law of Cosines for $\angle C$:$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$$\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(

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$1+\sqrt{3}$

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(B) Using the Law of Cosines for $\angle C$:$\cos(C) = \frac{a^2 + b^2 - c^2}{2ab}$$\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(x^2+x+1)(x^2-1)}$$\frac{\sqrt{3}}{2} = \frac{(x^4+x^2+1+2x^3+2x^2+2x) + (x^4-2x^2+1) - (4x^2+4x+1)}{2(x^2+x+1)(x^2-1)}$$\frac{\sqrt{3}}{2} = \frac{2x^4+2x^3-3x^2-2x+1}{2(x^2+x+1)(x^2-1)}$Since $(x^2+x+1)(x^2-1) = x^4+x^3-x-1$,the equation simplifies to:$\sqrt{3}(x^4+x^3-x-1) = 2x^4+2x^3-3x^2-2x+1$Solving this quadratic in $x$ leads to $x = 1+\sqrt{3}$ as the valid solution,since side lengths must be positive ($x > 1$ for $b > 0$).

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

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