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(D) Let the height of the tower $AB$ be $h$. Let the two points on the road be $C$ and $D$ such that $CD = 500 \ m$. Let $AD = x$. In $	riangle ABD$,

Vedclass · Accepted Answer

$250(3+\sqrt{3}) \ m$

Vedclass · Answer

(D) Let the height of the tower $AB$ be $h$. Let the two points on the road be $C$ and $D$ such that $CD = 500 \ m$. Let $AD = x$. In $	riangle ABD$,$	an 60^{\circ} = \frac{AB}{AD} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}$. In $	riangle ABC$,$	an 45^{\circ} = \frac{AB}{AC} = \frac{h}{x+500} \implies 1 = \frac{h}{\frac{h}{\sqrt{3}} + 500}$. $h = \frac{h}{\sqrt{3}} + 500 \implies h(1 - \frac{1}{\sqrt{3}}) = 500 \implies h(\frac{\sqrt{3}-1}{\sqrt{3}}) = 500$. $h = \frac{500 \sqrt{3}}{\sqrt{3}-1} \ m$. Rationalizing the denominator: $h = \frac{500 \sqrt{3}(\sqrt{3}+1)}{3-1} = \frac{500(3+\sqrt{3})}{2} = 250(3+\sqrt{3}) \ m$.

$A$ tower stands at the end of a straight road. The angles of elevation of the top of the tower from two points on the road $500 \ m$ apart are $45^{\circ}$ and $60^{\circ},$ respectively. Find the height of the tower.

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