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Question: In a triangle ABC, if the sides a, b, c are roots of x<sup>3</sup> – 11x<sup>2</sup> + 38x – 40 = 0...

In a triangle ABC, if the sides a, b, c are roots of

x3 – 11x2 + 38x – 40 = 0, then ∑cos⁡Aa\sum_{}^{}\frac{\cos A}{a}=

A

3/4

B

4/3

C

16/9

D

9/16

Answer

9/16

Explanation

Solution

Σa = 11, Σab = 38, abc = 40

Σcos⁡Aa\Sigma\frac{\cos A}{a} = 180Σ(b2+c2−a2)\frac{1}{80}\Sigma(b^{2} + c^{2} - a^{2})

= (Σa)2−2(Σab)80\frac{(\Sigma a)^{2} - 2(\Sigma ab)}{80}= 112−7680\frac{11^{2} - 76}{80}

= 4580\frac{45}{80} = 916\frac{9}{16}