+35 In triangle ABC, we know the following:
\(\begin{align*} \angle A &= 120^{\circ} \\ AB &= x \\ AC &= x + 2 \\ BC &= x + 4 \\ \end{align*}\)
Find \(x\).
+129852 We can use the Law of Cosines
BC^2 = AB^2 + AC^2 - 2 ( AB * AC) * cos (120)
(x + 4)^2 = x^2 + (x + 2)^2 - 2( x * ( x + 2)) (- 1/2) simplify
x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 + x^2 + 2x
x^2 + 8x + 16 = 3x^2 + 6x + 4
2x^2 - 2x - 12 = 0
x^2 -x - 6 = 0
(x -3) ( x + 2) = 0
X -3 = 0
x = 3
