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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`

A

`16/9`

B

`3/4`

C

`4/3`

D

`9/16`

Text Solution

Verified by Experts

We have,
a + b + c = 11, ab + bc + ca= 38 and abc = 40
`therefore(cosA)/a+(cosB)/b+(cosC)/c=(a^(2)+b^(2)+c^(2))/(2abc)`
`=((a+b+c)^(2)-2(ab+bc+ca))/(2abc)=(121-2xx38)/(2xx40)=9/16`
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