Question
In a $$\vartriangle ABC,$$ the sides $$a, b$$ and $$c$$ are such that they are the roots of $${x^3} - 11{x^2} + 38x - 40 = 0.$$ Then $$\frac{{\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{\cos C}}{c}$$ is equal to
A.
$$\frac{3}{4}$$
B.
$$1$$
C.
$$\frac{9}{16}$$
D.
None of these
Answer :
$$\frac{9}{16}$$
Solution :
Expression $$ = \frac{{{b^2} + {c^2} - {a^2}}}{{2abc}} + \frac{{{c^2} + {a^2} - {b^2}}}{{2abc}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2abc}} = \frac{{{a^2} + {b^2} + {c^2}}}{{2abc}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)}}{{2abc}} = \frac{{{{11}^2} - 2 \cdot 38}}{{2 \cdot 40}} = \frac{9}{{16}}.$$