A man on the top of a vertical tower observes a car moving at a...

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes

18min

for the angle of depression of the car to change from

30 °

45 °

, then the time taken (in

min

) by the car to reach the foot of the tower is A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road.

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road.

Options

A $\dfrac{9}{2}\left(\sqrt{3} - 1\right)$
B $18\left(1 + \sqrt{3}\right)$
C $18\left(\sqrt{3} - 1\right)$
D $9\left(1 + \sqrt{3}\right)$

Correct answer

D. $9\left(1 + \sqrt{3}\right)$

Step-by-step solution

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road.

Let length of tower be $h$ .

$⇒ A{C}^{′} = AB = h$

And $AC = ABcot30 ° = \sqrt{3}h$

$⇒ C{C}^{′} = \left(\sqrt{3} - 1\right)h$

Time taken by car form $C$ ${C}^{'} = 18min$

$⇒$ Time take by car to reach the foot of the tower

$= \dfrac{18}{\sqrt{3} - 1}min$

$= 9\left(\sqrt{3} + 1\right)min$

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