An aeroplane when 3000 m high passes vertically above another aeroplane at an instant when their angles of elevation at the same observing point are `60^(@)` and `45^(@)` respectively. How many metres higher is the one than the other?

Options

A. 1248 m
B. 1188 m
C. 1752 m
D. 1268 m

Answer

To solve the problem, we will use trigonometry, specifically the tangent function, which relates the angles of elevation to the heights of the planes and the distance from the observer to the point directly below the planes.

### Step-by-Step Solution:

1. **Identify the heights and angles**:
- Let the height of the first aeroplane (A) be \( h_A = 3000 \) m.
- The angle of elevation to the first aeroplane (A) is \( 60^\circ \).
- The angle of elevation to the second aeroplane (B) is \( 45^\circ \).

2. **Set up the relationships using tangent**:
- For the first aeroplane (A):
\[ \tan(60^\circ) = \frac{h_A}{d} \]
where \( d \) is the horizontal distance from the observer to the point directly below the first aeroplane.

- For the second aeroplane (B):
\[ \tan(45^\circ) = \frac{h_B}{d} \]
where \( h_B \) is the height of the second aeroplane.

3. **Calculate the horizontal distance \( d \)**:
- From the first equation:
\[ \tan(60^\circ) = \sqrt{3} \implies \sqrt{3} = \frac{3000}{d} \implies d = \frac{3000}{\sqrt{3}} \approx 1732.05 \text{ m} \]

4. **Calculate the height of the second aeroplane (B)**:
- From the second equation:
\[ \tan(45^\circ) = 1 \implies 1 = \frac{h_B}{d} \implies h_B = d = 1732.05 \text{ m} \]

5. **Find the difference in height between the two aeroplanes**:
- The difference in height is:
\[ \text{Difference} = h_A - h_B = 3000 - 1732.05 \approx 1267.95 \text{ m} \]

6. **Round the result**:
- The difference in height can be approximated to \( 1268 \text{ m} \).

### Final Answer:
The first aeroplane is approximately **1268 meters** higher than the second aeroplane.

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