Question

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

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

Answer 1

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

As a,b and c are the roots of the equation, $x^3-11x^2+38x-40=0$
So, a+b+c = 11, ab+bc+ac = 38 and abc = 40.
Using cosine Rule in the given statement, we get-

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2\mathrm{abc}}=\frac{{(a+b+c)}^2-2(ab+bc+ca)}{2\mathrm{abc}}=\frac{{11}^2-76}{80}=\frac{45}{80}=\frac{9}{16}$