Question

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at the angle of depression of \( 30^\circ \), which is approaching the foot of the tower with a uniform speed. After 6 seconds the angle of depression of the car is found to be \( 60^\circ \). Find the time taken by the car to reach the foot of the tower from this point.

Hint:

To solve problems involving angles of depression, use the tangent function to relate the height of the tower and the distance of the car from the tower.

Answer 1

Let the height of the tower be \( h \) and the distance of the car from the foot of the tower at the initial moment be \( x \). From the angle of depression of \( 30^\circ \), we can write: \[ \tan(30^\circ) = \frac{h}{x}. \] We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so: \[ \frac{1}{\sqrt{3}} = \frac{h}{x} \quad \Rightarrow \quad x = \sqrt{3}h. \] After 6 seconds, the angle of depression becomes \( 60^\circ \). Let the new distance of the car from the foot of the tower be \( y \). From the angle of depression of \( 60^\circ \), we can write: \[ \tan(60^\circ) = \frac{h}{y}. \] We know that \( \tan(60^\circ) = \sqrt{3} \), so: \[ \sqrt{3} = \frac{h}{y} \quad \Rightarrow \quad y = \frac{h}{\sqrt{3}}. \] The car moves from \( x \) to \( y \) in 6 seconds. So, the speed of the car is: \[ \text{Speed} = \frac{x - y}{6} = \frac{\sqrt{3}h - \frac{h}{\sqrt{3}}}{6}. \] Simplifying: \[ \text{Speed} = \frac{h}{6} \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = \frac{h}{6} \times \frac{2}{\sqrt{3}} = \frac{h}{3\sqrt{3}}. \] Now, the time taken by the car to reach the foot of the tower is: \[ \text{Time} = \frac{x}{\text{Speed}} = \frac{\sqrt{3}h}{\frac{h}{3\sqrt{3}}} = 3 \times 3 = 9 \, \text{seconds}. \]
Conclusion: The time taken by the car to reach the foot of the tower from this point is 9 seconds.