An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10 s apart is `30^(@)` , what is the speed of the aircraft ?

Options

A. 10 .8 m `s^(-1)`
B. 1963 m `s^(-1)`
C. 108 m `s^(-1)`
D. 196.3 m `s^(-1)`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Geometry
We have an aircraft flying at a height \( h = 3400 \) m. The observer on the ground sees the aircraft at two positions 10 seconds apart, subtending an angle of \( 30^\circ \) between these two positions.

### Step 2: Set Up the Right Triangle
Let:
- Point A be the position of the observer on the ground.
- Point C be the position of the aircraft at the first observation.
- Point B be the position of the aircraft at the second observation after 10 seconds.

The height of the aircraft creates a right triangle with the ground. The angle \( \theta = 30^\circ \) is the angle subtended at point A by the two positions of the aircraft.

### Step 3: Use the Tangent Function
From the right triangle formed, we can use the tangent function to find the horizontal distance \( d \) covered by the aircraft between the two positions.

The tangent of the angle is given by:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \]
Where:
- \( h = 3400 \) m (height of the aircraft)
- \( d \) is the horizontal distance from point A to the vertical line of the aircraft.

For \( \theta = 30^\circ \):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Thus, we can express \( d \) as:
\[ d = \frac{h}{\tan(30^\circ)} = 3400 \times \sqrt{3} \]

### Step 4: Calculate the Horizontal Distance
Now we calculate \( d \):
\[ d = 3400 \times \sqrt{3} \approx 3400 \times 1.732 \approx 5885.8 \text{ m} \]

### Step 5: Calculate the Speed of the Aircraft
The aircraft covers this distance \( d \) in \( t = 10 \) seconds. The speed \( V \) of the aircraft is given by:
\[ V = \frac{d}{t} = \frac{5885.8 \text{ m}}{10 \text{ s}} = 588.58 \text{ m/s} \]

### Step 6: Final Calculation
Now, we can finalize the speed:
\[ V \approx 588.58 \text{ m/s} \]

### Conclusion
The speed of the aircraft is approximately \( 588.58 \text{ m/s} \).

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