In a triangle ABC,with usual notations,the si | vedclass

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(A) The sides $a, b, c$ are the roots of the cubic equation $x^3-11x^2+38x-40=0$.By Vieta's formulas,we have:$a+b+c = 11$$ab+bc+ca = 38$$abc = 40$Usin

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$\frac{9}{16}$

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(A) The sides $a, b, c$ are the roots of the cubic equation $x^3-11x^2+38x-40=0$.By Vieta's formulas,we have:$a+b+c = 11$$ab+bc+ca = 38$$abc = 40$Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.Thus,$\frac{\cos A}{a} = \frac{b^2+c^2-a^2}{2abc}$.Similarly,$\frac{\cos B}{b} = \frac{a^2+c^2-b^2}{2abc}$ and $\frac{\cos C}{c} = \frac{a^2+b^2-c^2}{2abc}$.Adding these expressions:$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc} = \frac{a^2+b^2+c^2}{2abc}$.We know that $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 11^2 - 2(38) = 121 - 76 = 45$.Therefore,the sum is $\frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.

In a triangle $ABC$,with usual notations,the sides $a, b, c$ are the roots of the equation $x^3-11x^2+38x-40=0$. Then,find the value of $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$.

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