Step-by-step Solution:

1. Let \( h \) be the height of the tower. 2. Let \( D \) be the foot of the tower. 3. Let \( AD = x \), \( BD = y \), and \( CD = z \). Given that the angles of elevation from points \( A \), \( B \), and \( C \) are \( 30^\circ \), \( 45^\circ \), and \( 60^\circ \) respectively, we can write: \[ \tan 30^\circ = \frac{h}{x} \Rightarrow x = h \cot 30^\circ = h \sqrt{3} \] \[ \tan 45^\circ = \frac{h}{y} \Rightarrow y = h \cot 45^\circ = h \] \[ \tan 60^\circ = \frac{h}{z} \Rightarrow z = h \cot 60^\circ = \frac{h}{\sqrt{3}} \] 4. Given \( BC = 60 \, \text{m} \): Since \( B \) and \( C \) are collinear and on the same side of the tower: \[ y - z = 60 \] Substitute \( y \) and \( z \): \[ h - \frac{h}{\sqrt{3}} = 60 \] \[ h \left(1 - \frac{1}{\sqrt{3}}\right) = 60 \] \[ h \left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = 60 \] \[ h = 60 \cdot \frac{\sqrt{3}}{\sqrt{3} - 1} \] Rationalize the denominator: \[ h = 60 \cdot \frac{\sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 60 \cdot \frac{3 + \sqrt{3}}{3 - 1} = 60 \cdot \frac{3 + \sqrt{3}}{2} = 30 (3 + \sqrt{3}) = 90 + 30\sqrt{3} \] 5. Find \( AB \): Since \( A \) and \( B \) are collinear and on the same side of the tower: \[ AB = x - y = h \sqrt{3} - h = h (\sqrt{3} - 1) \] Substitute \( h \): \[ AB = (90 + 30\sqrt{3}) (\sqrt{3} - 1) \] Expand the product: \[ AB = 90\sqrt{3} - 90 + 90 - 30\sqrt{3} = 60\sqrt{3} \] Therefore, the length of \( AB \) is: \[ \boxed{60\sqrt{3} \, \text{m}} \]