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

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(B) Using the cosine rule 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

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$1+\sqrt{3}$

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(B) Using the cosine rule 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 $(x^2+x+1)(x^2-1) = x^4+x^3-x^2-x^2-x-1 = x^4+x^3-2x^2-x-1$,the numerator simplifies to $2(x^4+x^3-2x^2-x-1) + x^2+1$. Solving the equation leads to $x = 1+\sqrt{3}$. Since side lengths must be positive,$b = x^2-1 > 0 \implies x > 1$. Thus,$x = 1+\sqrt{3}$ is the valid solution.

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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