Question

In a triangle (ABC), with usual notations, the sides (a, b, c) are such that they are roots of the equation (x^3 - 11x^2 + 38x - 40 = 0) then (\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = )

• A. (\frac{9}{16})
• B. (\frac{3}{4})
• C. (1)
• D. [suspicious link removed]

Hint:

For any triangle, $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{a^2+b^2+c^2}{2abc}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The sides \(a, b, c\) of a triangle are roots of a cubic equation.
We need to find the value of the expression \(\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c}\).
Step 2: Key Formula or Approach:
Use Vieta's formulas for roots of \(x^3 + px^2 + qx + r = 0\):
Sum of roots \(a+b+c = -p\).
Sum of roots taken two at a time \(ab+bc+ca = q\).
Product of roots \(abc = -r\).
Use the cosine rule: \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\).
Step 3: Detailed Explanation:
From the equation \(x^3 - 11x^2 + 38x - 40 = 0\):
\(a + b + c = 11\)
\(ab + bc + ca = 38\)
\(abc = 40\)
Now, expand the target expression:
\[ \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc} \]
\[ = \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc} \]
\[ = \frac{a^2 + b^2 + c^2}{2abc} \]
We know \(a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+bc+ca)\).
Substituting the values:
\[ a^2 + b^2 + c^2 = (11)^2 - 2(38) = 121 - 76 = 45 \]
Now, substitute back into the expression:
\[ \text{Value} = \frac{45}{2 \times 40} = \frac{45}{80} = \frac{9}{16} \]
Step 4: Final Answer:
The required value is \(\frac{9}{16}\).