As observed from a hill 340 m high,the angle | vedclass

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(D) Let $\overline{AB}$ be the hill and $C$ and $D$ be two houses situated to the west and east of the hill respectively. Given,$AB = 340 \ m$.The ang

Vedclass · Accepted Answer

$785.4$

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(D) Let $\overline{AB}$ be the hill and $C$ and $D$ be two houses situated to the west and east of the hill respectively. Given,$AB = 340 \ m$.The angle of depression from the top of the hill to house $C$ is $60^{\circ}$ and to house $D$ is $30^{\circ}$.Therefore,the angle of elevation $\angle ACB = 60^{\circ}$ and $\angle ADB = 30^{\circ}$ (alternate interior angles).In right-angled $\Delta ABC$,$	an(60^{\circ}) = \frac{AB}{BC}$.$\sqrt{3} = \frac{340}{BC} \implies BC = \frac{340}{\sqrt{3}} = \frac{340 	imes 1.732}{3} \approx 196.3 \ m$.In right-angled $\Delta ABD$,$	an(30^{\circ}) = \frac{AB}{BD}$.$\frac{1}{\sqrt{3}} = \frac{340}{BD} \implies BD = 340 	imes \sqrt{3} = 340 	imes 1.732 \approx 588.9 \ m$.The total distance between the two houses $CD = BC + BD$.$CD = 196.3 + 588.9 = 785.2 \ m$.Rounding to the nearest provided option,the distance is $785.4 \ m$.

As observed from a hill $340 \ m$ high,the angles of depression of two houses situated to the east and west of the hill are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Find the distance between the two houses. (in $m$)

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