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In a triangle ABC, if the sides a,b,c, are roots of `x^3-11 x^2+38 x-40=0,` then find the value of `(cosA)/a+(cosB)/b+(cosC)/c`

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Here a, b, c are roots of equation `x^(3) - 11 x^(2) + 38 x - 40 = 0` Therefore,
`a + b + c = 11, ab + bc + ac = 38, and abc = 40`
`rArr (cos A)/(a) + (cos B)/(b) + (cos C)/(c) = (a^(2) + b^(2) + c^(2))/(2 abc)`
`= ((a + b + c)^(2) - 2 (ab + bc + ac))/(2abc)`
`= (11^(2) - 76)/(80) = (45)/(80) = (9)/(16)`
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