The angle of elevation of an aeroplane from a point on the ground is `45^@`. After a flight of 15 sec, the eleva-tion changes to `30^@`. If the aeroplane is flying at a height of 3000 meters, find the speed of the aeroplane.

Answer

To solve the problem step by step, we will use trigonometric principles and the information provided in the question.

### Step 1: Understand the Problem
We have an aeroplane flying at a height of 3000 meters. The angle of elevation from a point on the ground to the aeroplane changes from \(45^\circ\) to \(30^\circ\) after 15 seconds.

### Step 2: Draw a Diagram
Let's visualize the situation:
- Let point \(P\) be the position of the aeroplane when the angle of elevation is \(45^\circ\).
- Let point \(Q\) be the position of the aeroplane after 15 seconds when the angle of elevation is \(30^\circ\).
- Let point \(A\) be the point on the ground directly below the aeroplane at point \(P\).
- Let point \(B\) be the point on the ground directly below the aeroplane at point \(Q\).

### Step 3: Set Up the Trigonometric Relationships
1. **For the angle of elevation \(45^\circ\):**
\[ \tan(45^\circ) = \frac{\text{Height}}{\text{Distance from point A to point P}} \]
Since \(\tan(45^\circ) = 1\):
\[ 1 = \frac{3000}{d_1} \implies d_1 = 3000 \text{ meters} \]

2. **For the angle of elevation \(30^\circ\):**
\[ \tan(30^\circ) = \frac{\text{Height}}{\text{Distance from point A to point B}} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{3000}{d_2} \implies d_2 = 3000\sqrt{3} \text{ meters} \]

### Step 4: Calculate the Horizontal Distance Travelled
The horizontal distance travelled by the aeroplane in 15 seconds is the difference between \(d_2\) and \(d_1\):
\[ \text{Distance travelled} = d_2 - d_1 = 3000\sqrt{3} - 3000 \]
Factoring out \(3000\):
\[ \text{Distance travelled} = 3000(\sqrt{3} - 1) \text{ meters} \]

### Step 5: Calculate the Speed of the Aeroplane
Speed is defined as distance travelled over time taken. Here, the time taken is 15 seconds:
\[ \text{Speed} = \frac{\text{Distance travelled}}{\text{Time}} = \frac{3000(\sqrt{3} - 1)}{15} \]
Simplifying:
\[ \text{Speed} = 200(\sqrt{3} - 1) \text{ meters per second} \]

### Final Answer
The speed of the aeroplane is \(200(\sqrt{3} - 1)\) meters per second.

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