Let the height of tower be \(h\) and distances from tower base be \(d_A, d_B, d_C\) for points A, B, C respectively. From angles: \(\tan(30°) = \frac{h}{d_A}\) gives \(d_A = h\sqrt{3}\); \(\tan(45°) = \frac{h}{d_B}\) gives \(d_B = h\); \(\tan(60°) = \frac{h}{d_C}\) gives \(d_C = \frac{h}{\sqrt{3}}\). Since points are collinear in order A, B, C from farthest to nearest: \(AB = d_A - d_B = h\sqrt{3} - h = h(\sqrt{3}-1)\) and \(BC = d_B - d_C = h - \frac{h}{\sqrt{3}} = h(1-\frac{1}{\sqrt{3}}) = h\frac{\sqrt{3}-1}{\sqrt{3}}\). Ratio: \(\frac{AB}{BC} = \frac{h(\sqrt{3}-1)}{h\frac{\sqrt{3}-1}{\sqrt{3}}} = \sqrt{3}\). Therefore AB:BC = \(\sqrt{3}:1\).