Question

Por favor me ajudem (15pontos)

Determine o valor de x no triângulo abaixo:

Answer 1

Aplicando lei dos cossenos :

$\displaystyle \sf (x+2)^2=x^2+(x+1)^2-2\cdot x\cdot (x+1)\cdot cos(120^\circ) \\\\ Cos120^\circ =-cos(60^\circ) \\\\ Da{\'i}} \\\\ \sf (x+2)^2=x^2+(x+1)^2-2\cdot x\cdot (x+1)\cdot [-cos(60^\circ)] \\\\\ (x+2)^2=x^2+(x+1)^2+2\cdot x\cdot (x+1)\cdot cos(60^\circ)\\\\ x^2+4x+4=x^2+x^2+2x+1+2\cdot (x^2+x)\cdot \frac{1}{2} \\\\ x^2+4x+4=2x^2+2x+1+x^2+x \\\\ 3x^2-x^2+3x-4x+1-4= 0\\\\ 2x^2-x-3 = 0 \\\\ x = \frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 2\cdot(-3)}}{2\cdot 2}$

$\displaystyle \sf x = \frac{1\pm\sqrt{1+24}}{4} \to x=\frac{1\pm\sqrt{25}}{4} \\\\\\ x=\frac{1-5}{4}\to x=-1 (NAO\ CONV{\'EM}) \\\\\\\ x=\frac{1+5}{4} \to x= \frac{6}{4}=\frac{3}{2} \\\\\ Portanto : \\\\ \huge\boxed{\sf x=\frac{3}{2}}\checkmark$

OBS: segue a resolução normal caso esteja no celular e não carregue os códigos latex :

(x+2)²= x²+(x+1)²-2.x.(x+1).cos(120)

x²+4x+4=x²+x²+2x+1-2.(x²+x).(-1/2)

x²+4x+4=3x²+2x+1+x²+x

3x²-x²+3x²-4x+1-4 = 0

2x²-x²-3 = 0

Δ = (-1)²-4.2.(-3)

Δ = 1+24 = 24

Daí :

x = [-(-1)+√Δ]/2.2

x = [1+5]/4

x = 6/4

x = 3/2