Question

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 18 min. for the angle of depression of the car to change from $30^{\circ}$ to $45^{\circ}$ ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

• A. $9( 1 +\sqrt{3})$
• B. $ 18 ( 1 +\sqrt{3})$
• C. $18 ( \sqrt{3} - 1) $
• D. $\frac{9}{2} ( \sqrt{3} - 1) $

Answer 1

The Correct Option is A

To solve this problem, we need to understand the scenario involving angles of depression and the geometry associated with it. The problem is related to trigonometry, where a man at the top of a tower observes a car approaching the tower, and the angle of depression changes as the car moves towards the tower.

Let's denote:

• The height of the tower as h.
• The initial distance of the car from the tower when the angle of depression is 30^{\circ} as d_1.
• The distance of the car from the tower when the angle of depression is 45^{\circ} as d_2.

Using the tangent of angles in a right triangle, we have the following relationships:

For 30^{\circ}:

\tan 30^{\circ} = \frac{h}{d_1} \Rightarrow d_1 = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}

For 45^{\circ}:

\tan 45^{\circ} = \frac{h}{d_2} \Rightarrow d_2 = \frac{h}{\tan 45^{\circ}} = h

The car moves from d_1 to d_2 in 18 minutes. Therefore, the relative distance covered during this time is:

d_1 - d_2 = h\sqrt{3} - h = h(\sqrt{3} - 1)

The speed of the car is then calculated as:

\text{Speed} = \frac{h(\sqrt{3} - 1)}{18}

Now, we want to find the time taken by the car to reach the foot of the tower from d_2:

The remaining distance d_2 is h, and thus the time taken is given by:

t = \frac{d_2}{\text{Speed}} = \frac{h}{\frac{h(\sqrt{3} - 1)}{18}} = \frac{18}{\sqrt{3} - 1}

To simplify the calculation, we rationalize the denominator:

t = \frac{18}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{18(\sqrt{3} + 1)}{3 - 1} = 9(\sqrt{3} + 1)

Therefore, the time taken by the car to reach the foot of the tower is 9(1 + \sqrt{3}) minutes.