The sides a,b,c of a triangle ABC are the roots of `x^3 - 11x^2 +38x - 40 =0, " then " Sigma(cos A)/a =`

Options

A. `3/4`
B. 1
C. `9/16`
D. none

Answer

To solve the problem, we need to find the value of \( \Sigma \left( \frac{\cos A}{a} \right) \) where \( a, b, c \) are the roots of the polynomial equation \( x^3 - 11x^2 + 38x - 40 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial**:
The polynomial is given as \( x^3 - 11x^2 + 38x - 40 = 0 \). From Vieta's formulas, we can extract the following:
- \( a + b + c = 11 \) (coefficient of \( x^2 \) with a negative sign)
- \( ab + bc + ca = 38 \) (coefficient of \( x \))
- \( abc = 40 \) (constant term with a negative sign)

2. **Use the cosine rule**:
We know that:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
Similarly, we can express \( \cos B \) and \( \cos C \):
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

3. **Set up the summation**:
Now we can write:
\[ \Sigma \left( \frac{\cos A}{a} \right) = \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} \]
Substituting the expressions for \( \cos A, \cos B, \cos C \):
\[ \Sigma \left( \frac{\cos A}{a} \right) = \frac{b^2 + c^2 - a^2}{2abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{2abc} \]

4. **Combine the terms**:
Combine the terms in the numerator:
\[ = \frac{(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + (a^2 + b^2 - c^2)}{2abc} \]
Simplifying the numerator:
\[ = \frac{2a^2 + 2b^2 + 2c^2}{2abc} = \frac{a^2 + b^2 + c^2}{abc} \]

5. **Calculate \( a^2 + b^2 + c^2 \)**:
We know from the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
Substituting the known values:
\[ 11^2 = a^2 + b^2 + c^2 + 2 \cdot 38 \]
\[ 121 = a^2 + b^2 + c^2 + 76 \]
\[ a^2 + b^2 + c^2 = 121 - 76 = 45 \]

6. **Substitute back into the summation**:
Now substituting \( a^2 + b^2 + c^2 \) and \( abc \):
\[ \Sigma \left( \frac{\cos A}{a} \right) = \frac{45}{40} = \frac{9}{8} \]

### Final Answer:
Thus, the value of \( \Sigma \left( \frac{\cos A}{a} \right) \) is \( \frac{9}{8} \).