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