A man on the top of a vertical tower observes | vedclass

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Question

(A) Let the height of the tower be $h$ and the foot of the tower be $D$. Let the positions of the car be $B$ and $A$ at the angles of depression $30^\

Vedclass · Accepted Answer

$9(1 + \sqrt{3})$

Vedclass · Answer

(A) Let the height of the tower be $h$ and the foot of the tower be $D$. Let the positions of the car be $B$ and $A$ at the angles of depression $30^\circ$ and $45^\circ$ respectively.In $\Delta ODA$,$\angle OAD = 45^\circ$. Thus,$	an(45^\circ) = \frac{h}{DA} \Rightarrow DA = h$.In $\Delta ODB$,$\angle OBD = 30^\circ$. Thus,$	an(30^\circ) = \frac{h}{DB} \Rightarrow DB = h\sqrt{3}$.The distance covered by the car in $18 	ext{ min}$ is $BA = DB - DA = h(\sqrt{3} - 1)$.Speed of the car $v = \frac{	ext{distance}}{	ext{time}} = \frac{h(\sqrt{3} - 1)}{18}$.Time taken to cover distance $DA$ is $t = \frac{DA}{v} = \frac{h}{h(\sqrt{3} - 1) / 18} = \frac{18}{\sqrt{3} - 1}$.Rationalizing the denominator: $t = \frac{18(\sqrt{3} + 1)}{3 - 1} = \frac{18(\sqrt{3} + 1)}{2} = 9(\sqrt{3} + 1) 	ext{ min}$.

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