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(A) Given that $a, b, c$ are consecutive natural numbers with $a < b < c$,we have $a = b - 1$ and $c = b + 1$.Using the Law of Cosines,$\cos A = \frac

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$3b(b^2 - 2)$

Vedclass · Answer

(A) Given that $a, b, c$ are consecutive natural numbers with $a Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Substituting these into the expression $(\cos A + \cos B + \cos C) 2abc$:$= (\frac{b^2 + c^2 - a^2}{2bc} + \frac{a^2 + c^2 - b^2}{2ac} + \frac{a^2 + b^2 - c^2}{2ab}) 2abc$$= a(b^2 + c^2 - a^2) + b(a^2 + c^2 - b^2) + c(a^2 + b^2 - c^2)$Substitute $a = b - 1$ and $c = b + 1$:$= (b - 1)(b^2 + (b + 1)^2 - (b - 1)^2) + b((b - 1)^2 + (b + 1)^2 - b^2) + (b + 1)((b - 1)^2 + b^2 - (b + 1)^2)$$= (b - 1)(b^2 + 4b) + b(b^2 - 2b + 1 + b^2 + 2b + 1 - b^2) + (b + 1)(b^2 - 2b + 1 + b^2 - b^2 - 2b - 1)$$= (b - 1)(b^2 + 4b) + b(b^2 + 2) + (b + 1)(b^2 - 4b)$$= (b^3 + 4b^2 - b^2 - 4b) + (b^3 + 2b) + (b^3 - 4b^2 + b^2 - 4b)$$= 3b^3 - 6b = 3b(b^2 - 2)$.

In a triangle,if the lengths of the sides $a, b,$ and $c$ are three consecutive natural numbers and $a < b < c$,then $(\cos A + \cos B + \cos C) 2abc = $

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