Question

If the sides a, b, c of a triangle abc are the roots of the equation x3 – 13x2 + 54x – 72 = 0, then the value of

Answer 1

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Answer 2

The question is:

If 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

$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$ is

$\boxed{\frac{61}{144}}$

Step-by-step explanation:

The given cubic equation:

$x^3-13x^2+54x-72=0$ .......... (1)

Let the expression in x be

$p(x)=x^3-13x^2+54x-72$

Putting x=3 in the above expression, we get

$p(3)=3^3-13\times 3^2+54\times 3-72$

$\implies p(3)=27-117+162-72=0$

Thus x-3 is a factor of p(x)

or, x = 3 is a root of the given cubic equation (1)

Dividing p(x) by x-3 we can write it is

$p(x)=(x-3)(x^2-10x+24)$

$\implies p(x)=(x-3)(x^2-6x-4x+24)$

$\implies p(x)=(x-3)(x-4)(x-6)$

Therefore, the roots of the given cubic polynomial (1) are 3, 4, 6

From the solution of triangles we know that

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

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

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

Thus,

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

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

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

$=\frac{3^2+4^2+6^2}{2\times 3\times 4\times 6}$

$=\frac{9+16+36}{6\times 4\times 6}$

$=\frac{61}{144}$

Hope this answer is helpful.

Know More:

Q: If the sides of triangle ABC are in the ratio 4:5:6 prove that one angle is twice of the other.

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