Let the height of tower be \(h\) and distances from tower foot be \(x_A, x_B, x_C\). From the angles: \(\tan(30°) = \frac{h}{x_A}\) gives \(x_A = h\sqrt{3}\); \(\tan(45°) = \frac{h}{x_B}\) gives \(x_B = h\); \(\tan(60°) = \frac{h}{x_C}\) gives \(x_C = \frac{h}{\sqrt{3}}\). Since points are collinear with C closest to tower: \(AB = x_A - x_B = h\sqrt{3} - h = h(\sqrt{3}-1)\) and \(BC = x_B - x_C = h - \frac{h}{\sqrt{3}} = h\frac{\sqrt{3}-1}{\sqrt{3}}\). Therefore, \(\frac{AB}{BC} = \frac{h(\sqrt{3}-1)}{h\frac{\sqrt{3}-1}{\sqrt{3}}} = \sqrt{3}\). The ratio is \(\sqrt{3}:1\).