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(D) Let $h$ be the height of the tower $AB$ and $x$ be the length of the shadow when the Sun's altitude is $45^{\circ}$.In $	riangle ABD$,$	an 45^{\

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$30(\sqrt{3}+1) \ m$

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(D) Let $h$ be the height of the tower $AB$ and $x$ be the length of the shadow when the Sun's altitude is $45^{\circ}$.In $	riangle ABD$,$	an 45^{\circ} = \frac{AB}{AD} = \frac{h}{x}$. Since $	an 45^{\circ} = 1$,we have $h = x$.In $	riangle ABC$,the shadow length is $AC = AD + DC = x + 60$. The Sun's altitude is $30^{\circ}$.$	an 30^{\circ} = \frac{AB}{AC} = \frac{h}{x + 60}$.Substituting $x = h$,we get $\frac{1}{\sqrt{3}} = \frac{h}{h + 60}$.$h + 60 = h\sqrt{3}$.$60 = h(\sqrt{3} - 1)$.$h = \frac{60}{\sqrt{3} - 1}$.Rationalizing the denominator: $h = \frac{60(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{60(\sqrt{3} + 1)}{3 - 1} = \frac{60(\sqrt{3} + 1)}{2} = 30(\sqrt{3} + 1) \ m$.

The shadow of a tower standing on a level ground is found to be $60 \ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is

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