If the sides of a triangle are $x^{2} + x + 1, x^{2} - 1$ and $2 x + 1$ , then the greatest angle is.

90^{\circ}

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135^{\circ}

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115^{\circ}

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120^{\circ}

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Solution

The correct option is D $120^{\circ}$
$a = x^{2} + x + 1$
$b = x^{2} - 1$
$c = 2 x + 1$
Side $a$ is greater $∴$ angle opp. to side is greater
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

$= \frac{{(x^{2} - 1)}^{2} + {(2 x + 1)}^{2} - {(x^{2} + x + 1)}^{2}}{2 (x^{2} - 1) (2 x + 1)}$

$= \frac{x^{4} + 1 - 2 x^{2} + {4 x}^{2} + 1 + 4 x - x^{4} - x^{2} - 1 - 2 x^{3} - 2 x - 2 x^{2}}{2 (x^{2} - 1) (2 x + 1)}$

$= \frac{- 2 x^{3} - x^{2} + 2 x + 1}{2 (x^{2} - 1) (2 x + 1)}$

$= \frac{- 2 x (x^{2} - 1) - 1 (x^{2} - 1)}{2 (x^{2} - 1) (2 x + 1)}$

$= \frac{- (2 x + 1) (x^{2} - 1)}{2 (x^{2} - 1) (2 x + 1)}$

$= \frac{- 1}{2}$
$c o s A = - c o s 6 0^{o} = c o s (18 0^{o} - 6 0^{o})$
$A = 12 0^{o}$

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The length of the sides of a triangle are $x, y$ and $\sqrt{x^{2} + y^{2} + x y}$ and The measure of the greatest angle is

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