Solution:

Given \(|\vec{A}|=3, |\vec{B}|=4, |\vec{C}|=5\) and \(\vec{A} + \vec{B} = \vec{C}\). Squaring both sides: \(|\vec{A} + \vec{B}|^2 = |\vec{C}|^2\). \(|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|cos\theta = |\vec{C}|^2\). Substituting values: \(3^2 + 4^2 + 2(3)(4)cos\theta = 5^2\) which gives \(9 + 16 + 24cos\theta = 25\). Thus, \(24cos\theta = 0\) and \(\theta = \pi/2\).