If 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 `(cosA)/(a)+(cosB)/(b)+(cosC)/(c )` is equal to :

Options

A. `(61)/(144)`
B. `(61)/(72)`
C. `(169)/(144)`
D. `(59)/(144)`

Answer

To solve the problem, we need to find the value of \(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\) given that the sides \(a\), \(b\), and \(c\) of triangle \(ABC\) are the roots of the polynomial equation \(x^3 - 13x^2 + 54x - 72 = 0\).

### Step 1: Find the roots of the polynomial
We start by identifying the roots of the polynomial equation \(x^3 - 13x^2 + 54x - 72 = 0\). We can use the Rational Root Theorem to test for possible rational roots.

Testing \(x = 2\):
\[ 2^3 - 13(2^2) + 54(2) - 72 = 8 - 52 + 108 - 72 = -16 + 108 - 72 = 20 \quad \text{(not a root)} \]

Testing \(x = 3\):
\[ 3^3 - 13(3^2) + 54(3) - 72 = 27 - 117 + 162 - 72 = 27 - 117 + 162 - 72 = 0 \quad \text{(is a root)} \]

Now that we have found \(x = 3\) as a root, we can perform synthetic division to factor the polynomial:
\[ \begin{array}{r|rrrr} 3 & 1 & -13 & 54 & -72 \\ & & 3 & -30 & 72 \\ \hline & 1 & -10 & 24 & 0 \\ \end{array} \]
This gives us \(x^2 - 10x + 24 = 0\).

### Step 2: Solve the quadratic equation
Next, we solve the quadratic equation \(x^2 - 10x + 24 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(24)}}{2(1)} = \frac{10 \pm \sqrt{100 - 96}}{2} = \frac{10 \pm 2}{2} \]
This gives us the roots:
\[ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{8}{2} = 4 \]
Thus, the roots of the polynomial are \(a = 3\), \(b = 4\), and \(c = 6\).

### Step 3: Calculate \(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\)
Using the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting \(a = 3\), \(b = 4\), and \(c = 6\):
\[ \cos A = \frac{4^2 + 6^2 - 3^2}{2 \cdot 4 \cdot 6} = \frac{16 + 36 - 9}{48} = \frac{43}{48} \]
\[ \cos B = \frac{3^2 + 6^2 - 4^2}{2 \cdot 3 \cdot 6} = \frac{9 + 36 - 16}{36} = \frac{29}{36} \]
\[ \cos C = \frac{3^2 + 4^2 - 6^2}{2 \cdot 3 \cdot 4} = \frac{9 + 16 - 36}{24} = \frac{-11}{24} \]

Now we calculate:
\[ \frac{\cos A}{a} = \frac{43/48}{3} = \frac{43}{144}, \quad \frac{\cos B}{b} = \frac{29/36}{4} = \frac{29}{144}, \quad \frac{\cos C}{c} = \frac{-11/24}{6} = \frac{-11}{144} \]

### Step 4: Combine the results
Now we combine these fractions:
\[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{43}{144} + \frac{29}{144} - \frac{11}{144} = \frac{43 + 29 - 11}{144} = \frac{61}{144} \]

### Final Answer
Thus, the value of \(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\) is:
\[ \boxed{\frac{61}{144}} \]