An aeroplane flying horiontally 1 km above the ground is observed at an elevation of `60^@` and after 10 seconds the elevation is observed to be `30^@`. The uniform speed of the aeroplane in km/h is

Options

A. 240
B. `240sqrt3`
C. `60sqrt3`
D. None of these

Answer

To solve the problem of finding the uniform speed of the aeroplane, we will follow these steps:

### Step 1: Understand the scenario
The aeroplane is flying horizontally at a height of 1 km above the ground. It is observed at two different angles of elevation: first at \(60^\circ\) and then at \(30^\circ\) after 10 seconds.

### Step 2: Draw a diagram
Let's denote the points:
- Point A: The position of the aeroplane when the angle of elevation is \(60^\circ\).
- Point B: The position of the aeroplane when the angle of elevation is \(30^\circ\).
- Point O: The point on the ground directly below the aeroplane.

### Step 3: Set up the triangles
1. In triangle AOC (where angle AOC is \(30^\circ\)):
- The height (AC) is 1 km.
- We can use the tangent function:
\[ \tan(30^\circ) = \frac{AC}{OC} \]
- Thus, \(OC = \frac{AC}{\tan(30^\circ)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \text{ km}\).

2. In triangle BOD (where angle BOD is \(60^\circ\)):
- The height (BD) is again 1 km.
- Using the tangent function:
\[ \tan(60^\circ) = \frac{BD}{OD} \]
- Therefore, \(OD = \frac{BD}{\tan(60^\circ)} = \frac{1}{\sqrt{3}} \text{ km}\).

### Step 4: Calculate the distance traveled by the aeroplane
The distance \(CD\) that the aeroplane traveled in 10 seconds is given by:
\[ CD = OC - OD = \sqrt{3} - \frac{1}{\sqrt{3}} \]
To simplify:
\[ CD = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \text{ km} \]

### Step 5: Convert time from seconds to hours
The time taken is 10 seconds. To convert this into hours:
\[ \text{Time in hours} = \frac{10}{3600} = \frac{1}{360} \text{ hours} \]

### Step 6: Calculate the speed of the aeroplane
Using the formula for speed:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{\frac{2}{\sqrt{3}}}{\frac{1}{360}} = \frac{2 \times 360}{\sqrt{3}} = \frac{720}{\sqrt{3}} \text{ km/h} \]
To simplify:
\[ \text{Speed} = 240\sqrt{3} \text{ km/h} \]

### Final Answer
The uniform speed of the aeroplane is \(240\sqrt{3} \text{ km/h}\).

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