Question

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

• A. $1$
• B. $\frac{3}{4}$
• C. $\frac{9}{16}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{9}{16}$

Let, $k=\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}=\frac{2(a^2+b^2+c^2)-(a^2+b^2+c^2)}{2abc}$

$=\frac{a^2+b^2+c^2}{2abc}$

and $x^3-11x^2+38x-40=0$ has $a,b,c$ as roots.

$\Rightarrow a+b+c=11,$

$ab+bc+ac=38,$ and

$abc=40$

$\therefore k=\frac{{(a+b+c)}^2-2(ab+bc+ac)}{\left(2abc\right)}$

$=\frac{121-2\times 38}{2\times 40}$

$=\frac{45}{2\times 40}$

$\Rightarrow k=\frac{9}{16}$