Question

lf 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. $\frac{169}{144}$
• B. $\frac{61}{72}$
• C. $\frac{61}{144}$
• D. $\frac{169}{72}$

Answer 1

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

By Hit and Trial Method , we can find that

on putting $x=3$ we get

$\Rightarrow 27-117+162-72=0$

Therefore , 3 is the root of the Equation,

so we can write the Equation as

$\to x^2(x-3)-10x(x-3)+24(x-3)=(x-3)(x^2-10x+24)|\Rightarrow (x-3)(x-4)(x-6)$

Therefore, $a=3,b=6,c=4$

Using Cosine Formula

$cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{43}{48}$

Similarly, $cosB=\frac{-11}{24}cosC=\frac{29}{36}$

Now, $\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}=\frac{43}{144}-\frac{-11}{144}+\frac{29}{144}=\frac{61}{144}$

therefore, Correct Answer is $C$