Question

If the sides a, b, c of a triangle ABC are the roots of the equation $x^3-13x^2+54x-72=0$, then the value of $\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ is equal to

• A. $\frac{59}{144}$
• B. $\frac{61}{144}$
• C. $\frac{61}{72}$
• D. $\frac{32}{5}$

Answer 1

The correct option is B $\frac{61}{144}$

From equation
a + b + c = 13
ab + bc + ca = 54
abc = 72
$\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$
$=\frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$
$=\frac{(a+b+c{)}^2-2(ab+bc+ca)}{2abc}=\frac{61}{144}$