Question

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

• A. \(\frac{9}{16}\)
• B. \(\frac{3}{4}\)
• C. 1
• D. \(\frac{5}{16}\)

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: Simplify the expression
$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc \cdot a} + \dots$ is complex. Use Sine Rule instead:
$\frac{\cos A}{2R \sin A} = \frac{\cot A}{2R}$. Expression $= \frac{\cot A + \cot B + \cot C}{2R}$.
Alternatively, use $\frac{\cos A}{a} + \dots = \frac{(b^2+c^2-a^2)/2bc}{a} = \frac{b^2+c^2-a^2}{2abc}$.
Sum $= \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc} = \frac{a^2+b^2+c^2}{2abc}$.

Step 2: Use Theory of Equations
Sum of roots $a+b+c = 11$.
Sum of roots taken two at a time $ab+bc+ca = 38$.
Product of roots $abc = 40$.

Step 3: Calculate $a^2+b^2+c^2$
$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) = 11^2 - 2(38) = 121 - 76 = 45$.

Step 4: Final calculation
Value $= \frac{45}{2(40)} = \frac{45}{80} = \frac{9}{16}$.
Final Answer: (A)