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(B) Let $AB$ be the tower of height $h$ and $C$ be the initial position of the car. After $6$ seconds,the car reaches point $D$.In $	riangle ADB$,$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $AB$ be the tower of height $h$ and $C$ be the initial position of the car. After $6$ seconds,the car reaches point $D$.In $	riangle ADB$,$	an 60^{\circ} = \frac{AB}{DB} \implies \sqrt{3} = \frac{h}{DB} \implies DB = \frac{h}{\sqrt{3}}$.In $	riangle ABC$,$	an 30^{\circ} = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{h}{BC} \implies BC = h\sqrt{3}$.The distance covered in $6$ seconds is $CD = BC - DB = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$.Since the speed is uniform,the time taken to cover distance $CD$ is $6$ seconds.Therefore,time taken to cover distance $DB = \frac{h}{\sqrt{3}}$ is $\frac{6 	imes (h/\sqrt{3})}{2h/\sqrt{3}} = \frac{6}{2} = 3$ seconds.

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