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(B) Let the height of the building be $h \, m$ and the building be represented by $PQ$. Let the initial position of the person be $R$ and the position

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) Let the height of the building be $h \, m$ and the building be represented by $PQ$. Let the initial position of the person be $R$ and the position after $2 \, hours$ be $S$.Distance covered $RS = 	ext{speed} 	imes 	ext{time} = 25(\sqrt{3} - 1) 	imes 2 = 50(\sqrt{3} - 1) \, m$.In $\Delta PQS$,$	an(45^\circ) = \frac{PQ}{SQ} \implies 1 = \frac{h}{SQ} \implies SQ = h$.In $\Delta PQR$,$	an(30^\circ) = \frac{PQ}{RQ} = \frac{PQ}{RS + SQ} = \frac{h}{50(\sqrt{3} - 1) + h}$.Since $	an(30^\circ) = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{h}{50(\sqrt{3} - 1) + h}$.$\sqrt{3}h = 50(\sqrt{3} - 1) + h$.$(\sqrt{3} - 1)h = 50(\sqrt{3} - 1)$.$h = 50 \, m$.

$A$ person observes the angle of elevation of a building as $30^\circ$. The person proceeds towards the building with a speed of $25(\sqrt{3} - 1) \, m/hour$. After $2 \, hours$,he observes the angle of elevation as $45^\circ$. The height of the building (in meters) is:

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