Question

Let a, b and c be the sides of a $△ABC$. If $a^2,b^{2}and{c}^2$ are the roots of the equation $x^3-{Px}^2+Qx-R=0$, where P,Q & R are constants, then find the value of $\frac{cosA}{a}+\frac{cosB}{b}+\frac{cosC}{c}$ in terms of P,Q and R.

Answer 1

Given that a,b,c are lengths of $△$ ABC and $a^2$,$b^2$, $c^2$ are the roots of the cubic equation $x^3-P{x}^2+Qx-R=0$.

In a cubic equation, $a{x}^3+b{x}^2+cx+d=0$

Sum of roots can be calculated using the formula,$\frac{-b}{a}$ and

Product of the roots can be calculated using the formula,$\frac{-d}{a}$

So in the given cubic equation,

Sum of roots=$a^2+b^2+c^2=P$.

Product of Roots=$a^2{b}^2{c}^2=R$

In a triangle ABC,using the Cosine Rule we get

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos B=\frac{a^2+c^2-a^2}{2ac}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

On squaring them we get

$\Rightarrow \frac{\cos A}{a}=\frac{b^2+c^2-a^2}{2abc}$ $\to$(1)

$\Rightarrow \frac{\cos B}{b}=\frac{a^2+c^2-b^2}{2abc}$ $\to$(2)

$\Rightarrow \frac{\cos C}{c}=\frac{a^2+b^2-c^2}{2abc}$ $\to$(3)

Adding equations 1,2 and 3 we get,

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^2+b^2+c^2}{2abc}$

$=\frac{P}{\sqrt{2R}}$

This is the required answer.