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(D) Let the height of the tower be $h$ and the distance of point $A$ from the base of the tower $(B)$ be $x$.In $	riangle ABD$, $	an 60^{\circ} = \f

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$30 \sqrt{6}$

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(D) Let the height of the tower be $h$ and the distance of point $A$ from the base of the tower $(B)$ be $x$.In $	riangle ABD$, $	an 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.Now, the person moves $60 \ m$ perpendicular to $AB$ to point $C$. Thus, $AC = 60 \ m$ and $	riangle ABC$ is a right-angled triangle at $A$.The distance from point $C$ to the base $B$ is $BC = \sqrt{AC^2 + AB^2} = \sqrt{60^2 + x^2} = \sqrt{3600 + x^2}$.In $	riangle CBD$, the angle of elevation is $45^{\circ}$, so $	an 45^{\circ} = \frac{h}{BC} \Rightarrow 1 = \frac{h}{\sqrt{3600 + x^2}}$.Therefore, $h^2 = 3600 + x^2$.Substituting $x^2 = \frac{h^2}{3}$ into the equation: $h^2 = 3600 + \frac{h^2}{3}$.$h^2 - \frac{h^2}{3} = 3600 \Rightarrow \frac{2h^2}{3} = 3600$.$h^2 = 1800 	imes 3 = 5400$.$h = \sqrt{5400} = \sqrt{900 	imes 6} = 30 \sqrt{6} \ m$.

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