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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} = \fr

Vedclass · Accepted Answer

$30 \sqrt{6}$

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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}$ $\Rightarrow h = x \sqrt{3}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.The person moves $60 \ m$ perpendicular to $AB$ to reach point $C$. Thus,$AC = 60 \ m$ and $\angle CAB = 90^{\circ}$.In $	riangle ABC$,the distance from $C$ to the base $B$ is $CB = \sqrt{AC^2 + AB^2} = \sqrt{60^2 + x^2}$.In $	riangle CBD$,the angle of elevation is $45^{\circ}$,so $	an 45^{\circ} = \frac{h}{CB} = 1$.Therefore,$h = CB = \sqrt{3600 + x^2} \Rightarrow h^2 = 3600 + x^2$.Substituting $x^2 = \frac{h^2}{3}$ into the equation: $h^2 = 3600 + \frac{h^2}{3}$.$\frac{2h^2}{3} = 3600 \Rightarrow h^2 = 1800 	imes 3 = 5400$.$h = \sqrt{5400} = \sqrt{900 	imes 6} = 30 \sqrt{6} \ m$.

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