Question

An Air Traffic Controller is watching from the control tower a plane which was moving from the tower. The plane makes the angle of depression of 60 degree with ATC's eye when at the distance of 100 meters from the tower. After 20 seconds the angle of depression becomes 45 degree. Calculate the speed of the moving plane.

• A. 13.176 km/hr
• B. 12.176 km/hr
• C. 13.200 km/hr
• D. 13.500 km/hr

Hint:

The "angle of depression" from a high point to an object is always equal to the "angle of elevation" from that object back to the high point (alternate interior angles). In many problems, it's easier to work with the angle of elevation from the ground.

Answer 1

The Correct Option is A

Step 1: Understanding the Problem
This is a problem involving trigonometry and the concept of speed, distance, and time. Due to the ambiguity of "angle of depression" to a moving plane, we interpret the angles as angles of elevation from the base of the tower to the plane, which maintains a constant altitude.
Let the constant height of the plane be \(h\).
Initial position (\(t=0\)): Horizontal distance from tower base (\(d_1\)) = 100 m. Angle of elevation = 60°.
Final position (\(t=20\)s): Horizontal distance (\(d_2\)). Angle of elevation = 45°.

Step 2: Key Formula or Approach
\begin{enumerate}
Use trigonometry (\(\tan\theta\)) to find the constant height (\(h\)) of the plane.
Use \(h\) to find the final horizontal distance (\(d_2\)).
Calculate the distance traveled by the plane (\(d_2 - d_1\)).
Calculate the speed using Speed = Distance / Time.
Convert the speed from m/s to km/hr. \end{enumerate}
Step 3: Detailed Explanation
1. Find the height of the plane (h):
From the initial position: \[ \tan(60^\circ) = \frac{\text{height}}{\text{distance}_1} = \frac{h}{100} \] \[ h = 100 \times \tan(60^\circ) = 100\sqrt{3} \text{ meters} \] 2. Find the final horizontal distance (\(d_2\)):
From the final position, the height is the same. \[ \tan(45^\circ) = \frac{\text{height}}{\text{distance}_2} = \frac{100\sqrt{3}}{d_2} \] Since \( \tan(45^\circ) = 1 \): \[ 1 = \frac{100\sqrt{3}}{d_2} \implies d_2 = 100\sqrt{3} \text{ meters} \] 3. Calculate the distance traveled:
The plane moved from a horizontal distance of \(d_1 = 100\) m to \(d_2 = 100\sqrt{3}\) m. \[ \text{Distance traveled} = d_2 - d_1 = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1) \text{ meters} \] Using \( \sqrt{3} \approx 1.732 \): \[ \text{Distance traveled} \approx 100(1.732 - 1) = 100(0.732) = 73.2 \text{ meters} \] 4. Calculate the speed in m/s:
Time taken = 20 seconds. \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{73.2 \text{ m}}{20 \text{ s}} = 3.66 \text{ m/s} \] 5. Convert speed to km/hr:
To convert m/s to km/hr, we multiply by \( \frac{18}{5} \). \[ \text{Speed in km/hr} = 3.66 \times \frac{18}{5} = \frac{65.88}{5} = 13.176 \text{ km/hr} \]
Step 4: Final Answer
The speed of the moving plane is 13.176 km/hr. Therefore, option (A) is the correct answer.