Question

In a triangle ABC , the sides \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are such that they are the roots of the equation \(x^3-11 x^2+38 x-40=0\) Then

\[ \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}= \]

• A. \(\frac{3}{4}\)
• B. 1
• C. \(\frac{9}{16}\)
• D. \(\frac{1}{16}\)

Hint:

Write each cosine with the cosine rule and use the sums of roots of the cubic.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Use \(\cos A=\frac{b^2+c^2-a^2}{2bc}\) so that \(\frac{\cos A}{a}=\frac{b^2+c^2-a^2}{2abc}\).

Step 2: Add the three terms
\(\sum\frac{\cos A}{a}=\frac{(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)}{2abc}=\frac{a^2+b^2+c^2}{2abc}\).

Step 3: Use the cubic
\(a+b+c=11\), \(ab+bc+ca=38\), \(abc=40\).
\(a^2+b^2+c^2=121-76=45\).

Step 4: Substitute
Value \(=\frac{45}{2\cdot40}=\frac{9}{16}\).

Final Answer:
The expression equals \(\frac{9}{16}\). \[ \boxed{\text{(C) }\dfrac{9}{16}} \]