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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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Answer» 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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