Question

The sides of a triangle are x^2+3x+3,2x+3,x^2+2x. prove that

Answer 1

Answer:

$120^{\circ}$

Step-by-step explanation:

$a=x^2+3x+3\\b=2x+3\\c=x^2+2x$

As a is the longest side, so A will be the largest angle

By Law of Cosines,

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

$\implies \cos A = \dfrac{(2x+3)^2+x^2(x+2)^2-(x^2+3x+3)^2}{2x(x+2)(2x+3)}$

$\implies \cos A=\dfrac{4x^2+12x+9+x^4+4x^3+4x^2-x^4-9x^2-9-6x^3-18x-6x^2}{2x(x+2)(2x+3)}$

$\implies \cos A=\dfrac{-2x^3-7x^2-6x}{2x(x+2)(2x+3)}$

$\implies \cos A=-\dfrac{2x^2+7x+6}{2(x+2)(2x+3)}$

$\implies \cos A = -\dfrac{(2x+3)(x+2)}{2(x+2)(2x+3)}$

$\implies \cos A=\dfrac{-1}{2}$

$\therefore A=120^{\circ}$