Question

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

• A. $\frac{9}{16}$
• B. $\frac{7}{8}$
• C. $\frac{9}{8}$
• D. $\frac{7}{16}$

Answer 1

The correct option is A $\frac{9}{16}$

Here a,b,c are roots of equation $x^3-11x^2+38x-40=0$

Therefore, $a+b+c=11,$,$ab+bc+ac=38,$ $abc=40,$

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

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

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

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

$=\frac{{11}^2-76}{80}$ $=\frac{45}{80}=\frac{9}{16}$