Question

If the sides $a,b,c$ are sides of a triangle $ABC$ are roots of the equation $x^3-15x^2+47x-82=0$ then the value of $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ equal to

• A. $\frac{225}{164}$
• B. $\frac{131}{164}$
• C. $\frac{131}{82}$
• D. $\frac{169}{82}$

Answer 1

The correct option is B $\frac{131}{164}$

$Equation{x}^3-15x^2+47x-82=0hasrootsa,b,c$

$Weknow,$

$a+b+c=15$

$ab+bc+ca=47$

$abc=82$

Now,

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$

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

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

$=\frac{{(a+b+c)}^2-2(ab+bc+ca)}{2abc}=\frac{225-2\times 47}{2\times 82}=\frac{131}{164}$