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Question

(A) Let the sides of the triangle be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$. Since $x^2+x+1$ is the greatest side,the greatest angle $A$ is opposite

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let the sides of the triangle be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$. Since $x^2+x+1$ is the greatest side,the greatest angle $A$ is opposite to side $a$. Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$. Substituting the values: $\cos A = \frac{(2x+1)^2 + (x^2-1)^2 - (x^2+x+1)^2}{2(2x+1)(x^2-1)}$. Expanding the terms: $(2x+1)^2 = 4x^2+4x+1$,$(x^2-1)^2 = x^4-2x^2+1$,and $(x^2+x+1)^2 = x^4+x^2+1+2x^3+2x^2+2x = x^4+2x^3+3x^2+2x+1$. Numerator: $(4x^2+4x+1) + (x^4-2x^2+1) - (x^4+2x^3+3x^2+2x+1) = -2x^3-x^2+2x+1$. Factoring the numerator: $-(2x^3+x^2-2x-1) = -[x^2(2x+1) - 1(2x+1)] = -(x^2-1)(2x+1)$. Thus,$\cos A = \frac{-(x^2-1)(2x+1)}{2(2x+1)(x^2-1)} = -\frac{1}{2}$. Therefore,$A = 120^{\circ}$.

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

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