Multiple choice

Let $a$, $b$ and $c$ be the sides of a $ \Delta ABC$. If $a^2, b^2 :and :c^2$ are the roots of the equation $\left (x^3 - Px^2 + Qx - R = 0\right )$, where P, Q and R are constants then the value of $ \left (\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}\right )$ is equal to

  1. $\dfrac{P}{2\sqrt{R}}$
  2. $\dfrac{P}{\sqrt{2R}}$
  3. $\dfrac{P\sqrt{2}}{R}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By Vieta's formulas for the given cubic equation, the sum of the roots is a^2 + b^2 + c^2 = P, and the product of the roots is a^2 * b^2 * c^2 = R, which gives abc = sqrt(R). Using the law of cosines, we can write cosA/a = (b^2 + c^2 - a^2) / (2abc). Summing this expression for all three angles simplifies to (a^2 + b^2 + c^2) / (2abc), which equals P / (2*sqrt(R)).

AI explanation

Using the projection formula, cos A = (b^2 + c^2 - a^2)/(2bc), we rewrite (cos A)/a as (b^2 + c^2 - a^2)/(2abc). By symmetry, we sum the expressions for all three angles to get (2(a^2 + b^2 + c^2) - (a^2 + b^2 + c^2))/(2abc) = (a^2 + b^2 + c^2)/(2abc). Since a^2, b^2, and c^2 are roots of x^3 - Px^2 + Qx - R = 0, their sum is P and their product is R, meaning (abc)^2 = R and the expression simplifies to P/(2R^(1/2)). The result is P/(2R^(1/2)).

Keep practicing — related questions