Question

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

• A. $1$
• B. $\frac{3}{4}$
• C. $\frac{9}{16}$
• D. None of these

Answer 1

Answer: (C)

$\frac{9}{16}$

Since $a,b,c$ are roots of $x^3-11x^2+38x-40=0$
$\therefore a+b+c=11,ab+bc+ca=38$
and $abc=40$
Using cosine rule in
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ we have
$=\frac{b^2+c^2-a^2}{2abc}+\frac{c^2+a^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$
$=\frac{a^2+b^2+c^2}{2abc}$
$=\frac{{(a+b+c)}^2-2(ab+bc+ca)}{2abc}$
$=\frac{{11}^2-2\times 38}{2\times 40}$
$=\frac{121-76}{80}$
$=\frac{45}{80}=\frac{9}{16}$