The greatest angle of the triangle whose side | vedclass

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(D) Let the sides be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$. For $x > 1$,$x^2+x+1$ is the largest side.Let $	heta$ be the angle opposite to the sid

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Let the sides be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$. For $x > 1$,$x^2+x+1$ is the largest side.Let $	heta$ be the angle opposite to the side $a = x^2+x+1$.Using the Law of Cosines: $\cos 	heta = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting the values:$\cos 	heta = \frac{(2x+1)^2 + (x^2-1)^2 - (x^2+x+1)^2}{2(2x+1)(x^2-1)}$$= \frac{(4x^2+4x+1) + (x^4-2x^2+1) - (x^4+x^2+1+2x^3+2x^2+2x)}{2(2x+1)(x^2-1)}$$= \frac{x^4+2x^2+4x+2 - (x^4+2x^3+3x^2+2x+1)}{2(2x+1)(x^2-1)}$$= \frac{-2x^3-x^2+2x+1}{2(2x+1)(x^2-1)}$$= \frac{-x^2(2x+1) + 1(2x+1)}{2(2x+1)(x^2-1)}$$= \frac{(2x+1)(1-x^2)}{2(2x+1)(x^2-1)}$$= \frac{-(x^2-1)}{2(x^2-1)} = -\frac{1}{2}$.Since $\cos 	heta = -\frac{1}{2}$,we have $	heta = 120^{\circ}$.

The greatest angle of the triangle whose sides are $x^2+x+1$,$2x+1$,and $x^2-1$ is (in $^{\circ}$)

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