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`

Explanation

As `a,b and c` are the roots of the equation,

`x^3-11x^2+38x-40 = 0`

`:. a+b+c = 11`

`ab+bc+ca = 38`

`abc = 40`

Now, `cosA/a+cosB/b+cosC/c = (b^2+c^2-a^2)/(2abc)+(a^2+c^2-b^2)/(2abc)+(a^2+b^2-c^2)/(2abc)`

`=(a^2+b^2+c^2)/(2abc)`

`=((a+b+c)^2-2(ab+bc+ca))/(2abc)`

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