If the sides of a Delta are (x^2 + x + 1),(2x | vedclass

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(B) Let the sides be $a = x^2 + x + 1$,$b = 2x + 1$,and $c = x^2 - 1$.The greatest side subtends the greatest angle. Comparing the sides for $x > 1$,$

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Let the sides be $a = x^2 + x + 1$,$b = 2x + 1$,and $c = x^2 - 1$.The greatest side subtends the greatest angle. Comparing the sides for $x > 1$,$x^2 + x + 1$ is the largest.Using the Law of Cosines,$\cos 	heta = \frac{b^2 + c^2 - a^2}{2bc}$.$\cos 	heta = \frac{(2x + 1)^2 + (x^2 - 1)^2 - (x^2 + x + 1)^2}{2(2x + 1)(x^2 - 1)}$.Expanding the terms:$(2x + 1)^2 = 4x^2 + 4x + 1$$(x^2 - 1)^2 = x^4 - 2x^2 + 1$$(x^2 + x + 1)^2 = x^4 + x^2 + 1 + 2x^3 + 2x^2 + 2x = x^4 + 2x^3 + 3x^2 + 2x + 1$.Numerator $= (4x^2 + 4x + 1) + (x^4 - 2x^2 + 1) - (x^4 + 2x^3 + 3x^2 + 2x + 1) = -2x^3 - x^2 + 2x + 1$.Factorizing the numerator: $-(2x^3 + x^2 - 2x - 1) = -[x^2(2x + 1) - 1(2x + 1)] = -(x^2 - 1)(2x + 1)$.$\cos 	heta = \frac{-(x^2 - 1)(2x + 1)}{2(2x + 1)(x^2 - 1)} = -\frac{1}{2}$.Since $\cos 	heta = -\frac{1}{2}$,$	heta = 120^o$.

If the sides of a $\Delta$ are $(x^2 + x + 1)$,$(2x + 1)$,and $(x^2 - 1)$,then the greatest angle is.....$^o$

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