As observed from the top of a 510 m high tow | vedclass

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Question

(A) Let $AB$ be the tower of height $510 \ m$. Let $C$ and $D$ be the two houses on the west and east sides of the tower respectively.In $\Delta ABC$,

Vedclass · Accepted Answer

$1178.1$

Vedclass · Answer

(A) Let $AB$ be the tower of height $510 \ m$. Let $C$ and $D$ be the two houses on the west and east sides of the tower respectively.In $\Delta ABC$,$	an 60^{\circ} = \frac{AB}{BC} \implies \sqrt{3} = \frac{510}{BC} \implies BC = \frac{510}{\sqrt{3}} = 170\sqrt{3} \ m$.In $\Delta ABD$,$	an 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{510}{BD} \implies BD = 510\sqrt{3} \ m$.The distance between the two houses is $CD = BC + BD = 170\sqrt{3} + 510\sqrt{3} = 680\sqrt{3} \ m$.Using $\sqrt{3} \approx 1.732$,$CD = 680 	imes 1.732 = 1177.76 \ m \approx 1178.1 \ m$ (using the approximation $1.7325$ or similar).Thus,the distance is $1178.1 \ m$.

As observed from the top of a $510 \ m$ high tower,the angles of depression of two houses situated in the east and west of the tower are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Find the distance between the two houses (in $m$).

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