Multiple choice

lf the sides $a,\ b,\ c$ of a triangle $ABC$ are the roots of the equation $x^{3}-13x^{2}+54x-72=0$, then the value of $\displaystyle \frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ is equal to?

  1. $\displaystyle \frac{169}{144}$
  2. $\displaystyle \frac{61}{72}$
  3. $\displaystyle \frac{61}{144}$
  4. $\displaystyle \frac{169}{72}$
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C Correct answer
Explanation

Let the roots be a, b, c. From Vieta's formulas, a+b+c = 13, ab+bc+ca = 54, abc = 72. Using the cosine rule, cosA = (b^2+c^2-a^2)/(2bc). The expression (cosA/a + cosB/b + cosC/c) simplifies to (a^2+b^2+c^2)/(2abc). Using (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca), we get 169 = a^2+b^2+c^2 + 108, so a^2+b^2+c^2 = 61. The result is 61 / (2 * 72) = 61/144.

AI explanation

Using Vieta's formulas for the cubic equation, the sum of the roots a + b + c = 13, the sum of their products ab + bc + ca = 54, and their product abc = 72. Apply the cosine formula to rewrite the expression as (b^2 + c^2 - a^2)/2abc + (a^2 + c^2 - b^2)/2abc + (a^2 + b^2 - c^2)/2abc. Adding the numerators yields (a^2 + b^2 + c^2)/(2abc), which can be rearranged using the symmetric sums as ((a + b + c)^2 - 2(ab + bc + ca))/(2abc). Substitute the values into this expression to calculate (169 - 108)/(2 * 72), which simplifies to 61/144.

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