Question

A tower stands vertically on the ground. A man standing at the top of the tower observes his friend at an angle of depression of 30\(^{\circ}\), who is approaching the foot of the tower with a uniform speed. 30 seconds later, the angle of depression changes to 60\(^{\circ}\). Find the time taken by his friend to reach the foot of the tower from this point.

Hint:

For any motion problem with angles changing from \( 30^{\circ} \) to \( 60^{\circ} \) towards the base:
The time taken to cover the remaining distance is always exactly half the time taken to cover the first interval.
Since the first interval took 30 seconds, the second interval takes exactly \( 30 / 2 = 15 \) seconds.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Heights and Distances (Trigonometry with Uniform Speed).
A person at the top of a tower observes an object on the ground. The initial angle of depression is \( 30^{\circ} \).
After 30 seconds of moving towards the tower at a uniform speed, the angle of depression becomes \( 60^{\circ} \).
We need to calculate the remaining time required to reach the base of the tower.

Step 2: Key Formula or Approach:
- Let the height of the tower be \( h \).
- Let \( C \) be the foot of the tower, and \( D \) be the top.
- Let the initial and final positions of the friend be \( A \) and \( B \) respectively.
- Set up trigonometric equations for \( \Delta DAC \) and \( \Delta DBC \).
- Relate distances to time using the uniform speed formula:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]

Step 3: Detailed Explanation:
1. Let \( h \) be the vertical height \( CD \) of the tower.
The angle of elevation of the top of the tower from the initial point \( A \) is \( 30^{\circ} \).
The angle of elevation from the second point \( B \) after 30 seconds is \( 60^{\circ} \).
2. In right-angled triangle \( \Delta DBC \) (with angle \( 60^{\circ} \) at \( B \)):
\[ \tan 60^{\circ} = \frac{CD}{BC} \]
\[ \sqrt{3} = \frac{h}{BC} \implies BC = \frac{h}{\sqrt{3}} \quad \text{(Equation 1)} \]
3. In right-angled triangle \( \Delta DAC \) (with angle \( 30^{\circ} \) at \( A \)):
\[ \tan 30^{\circ} = \frac{CD}{AC} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{AC} \implies AC = h\sqrt{3} \quad \text{(Equation 2)} \]
4. Calculate the distance \( AB \) covered by the friend in 30 seconds:
\[ AB = AC - BC \]
\[ AB = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\frac{3 - 1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}} \]
5. Let the uniform speed of the friend be \( v \). Since this distance was covered in 30 seconds:
\[ v = \frac{\text{Distance } AB}{\text{Time}} = \frac{2h / \sqrt{3}}{30} = \frac{h}{15\sqrt{3}}\text{ m/s} \]
6. Now, find the time \( t \) required to cover the remaining distance \( BC \):
\[ t = \frac{\text{Distance } BC}{v} \]
Substitute the expressions for \( BC \) and \( v \):
\[ t = \frac{h / \sqrt{3}}{h / (15\sqrt{3})} \]
Cancel common factors \( h \) and \( \sqrt{3} \):
\[ t = 15\text{ seconds} \]

Step 4: Final Answer:
The time taken by his friend to reach the foot of the tower from that point is 15 seconds.