Question

In a $\Delta ABC,$ the sides a,b and c are such that they are the roots of $x^3-11x^2+38x-40=0.$

Then $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ is equal to?

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

Answer 1

Answer: (C)

$\frac{9}{16}$

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

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

$=\frac{a^2+b^2+c^2}{2abc}$ ...(i)

Now, $x^3-11x^2+38x-40=0$

$(x-2)(x-4)(x-5)=0$

$x=2,4,5$

Hence, $a=2,b=4,c=5$.

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

$=\frac{4+16+25}{80}$

$=\frac{45}{80}$

$=\frac{9}{16}$.