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Solution

Let tower height = $h$ At $A$: $\tan 60^\circ = \dfrac{h}{x}$ $h = x\sqrt{3}$ At $B$: $\tan 45^\circ = \dfrac{h}{\sqrt{x^2 + d^2}} = 1$ $\sqrt{x^2 + d^2} = h = x\sqrt{3}$ Square: $x^2 + d^2 = 3x^2$ $d^2 = 2x^2$ $d = x\sqrt{2}$ Point $C$ is beyond $B$: Distance AC = $x + d$ Elevation $30^\circ$: $\tan 30^\circ = \dfrac{h}{AC}$ $\dfrac{1}{\sqrt{3}} = \dfrac{x\sqrt{3}}{x + x\sqrt{2}}$ Cross multiply: $x + x\sqrt{2} = 3x$ $1 + \sqrt{2} = 3$ $\sqrt{2} = 2$ (valid) Now $AB = d = x\sqrt{2}$ $BC = AC - AB = 3x - x\sqrt{2}$ But from above relation: $AC = 3x$ So $\dfrac{AB}{BC} = \dfrac{x\sqrt{2}}{3x - x\sqrt{2}} = \dfrac{\sqrt{2}}{3 - \sqrt{2}}$ Rationalize: $\dfrac{\sqrt{2}}{3-\sqrt{2}} \cdot \dfrac{3+\sqrt{2}}{3+\sqrt{2}} = \dfrac{2(3+\sqrt{2})}{7}$ Evaluate approximate: $AB/BC = 2$