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The greatest angle of the triangle whose sides are $x^2+x+1,2 x+1$ and $x^2-1$ is
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Verified Answer
The correct answer is:
$120^{\circ}$
Among $x^2+x+1,2 x+1, x^2-1$, the largest value is $x^2+x+1$.
$\therefore$ Greatest angle with the angle opposite to the side $x^2+x+1$. Let it be $\theta$.
$$
\begin{aligned}
\therefore \cos \theta & =\frac{(2 x+1)^2+\left(x^2-1\right)^2-\left(x^2+x+1\right)^2}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{4 x^2+4 x+1+x^4-2 x^2+1-x^4}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{-2 x^3-x^2+2 x+1}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{-x^2(2 x+1)+1(2 x+1)}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{(2 x+1)\left(1-x^2\right)}{2(2 x+1)\left(x^2-1\right)}=-\frac{1}{2} \\
\therefore \quad \theta & =120^{\circ}
\end{aligned}
$$
$\therefore$ Greatest angle with the angle opposite to the side $x^2+x+1$. Let it be $\theta$.
$$
\begin{aligned}
\therefore \cos \theta & =\frac{(2 x+1)^2+\left(x^2-1\right)^2-\left(x^2+x+1\right)^2}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{4 x^2+4 x+1+x^4-2 x^2+1-x^4}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{-2 x^3-x^2+2 x+1}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{-x^2(2 x+1)+1(2 x+1)}{2(2 x+1)\left(x^2-1\right)} \\
& =\frac{(2 x+1)\left(1-x^2\right)}{2(2 x+1)\left(x^2-1\right)}=-\frac{1}{2} \\
\therefore \quad \theta & =120^{\circ}
\end{aligned}
$$
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