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(C) Let $H$ be the height of the tower and $x$ be the distance between the building and the tower.From the top of the building (height $h = 60\sqrt{3}

Vedclass · Accepted Answer

$90(\sqrt{3}+1)$

Vedclass · Answer

(C) Let $H$ be the height of the tower and $x$ be the distance between the building and the tower.From the top of the building (height $h = 60\sqrt{3}$ feet),the angle of elevation to the top of the tower is $45^{\circ}$. Thus,$	an 45^{\circ} = \frac{H - h}{x} \implies 1 = \frac{H - 60\sqrt{3}}{x} \implies x = H - 60\sqrt{3}$.From the bottom of the building,the angle of elevation to the top of the tower is $60^{\circ}$. Thus,$	an 60^{\circ} = \frac{H}{x} \implies \sqrt{3} = \frac{H}{x} \implies x = \frac{H}{\sqrt{3}}$.Equating the two expressions for $x$: $H - 60\sqrt{3} = \frac{H}{\sqrt{3}}$.Multiply by $\sqrt{3}$: $\sqrt{3}H - 60(3) = H \implies \sqrt{3}H - H = 180 \implies H(\sqrt{3} - 1) = 180$.$H = \frac{180}{\sqrt{3} - 1} = \frac{180(\sqrt{3} + 1)}{3 - 1} = \frac{180(\sqrt{3} + 1)}{2} = 90(\sqrt{3} + 1)$ feet.

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