The angles of elevation of an aeroplane flying vertically above the ground , as observed from the two consecutive stones , 1 km apart , are `45^(@)and60^(@)` aeroplane from the ground is :

Options

A. `(sqrt(3)+1)` km .
B. `(sqrt(3)+3)` km .
C. `(1)/(2)(sqrt(3)+1)` km .
D. `(1)/(2)(sqrt(3)+3)` km .

Answer

To find the height of the aeroplane from the ground based on the angles of elevation observed from two consecutive stones, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- Let the height of the aeroplane be \( h \).
- Let the distance from the first stone to the point directly below the aeroplane be \( x \).
- The distance between the two stones is given as 1 km.

2. **Set Up the Right Triangles**:
- From the first stone, where the angle of elevation is \( 45^\circ \):
\[ \tan(45^\circ) = \frac{h}{x} \]
Since \( \tan(45^\circ) = 1 \):
\[ h = x \quad \text{(Equation 1)} \]

- From the second stone, where the angle of elevation is \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{x + 1} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ h = \sqrt{3}(x + 1) \quad \text{(Equation 2)} \]

3. **Substitute Equation 1 into Equation 2**:
- From Equation 1, we know \( x = h \). Substitute this into Equation 2:
\[ h = \sqrt{3}(h + 1) \]

4. **Solve for \( h \)**:
- Expand the equation:
\[ h = \sqrt{3}h + \sqrt{3} \]
- Rearranging gives:
\[ h - \sqrt{3}h = \sqrt{3} \]
\[ h(1 - \sqrt{3}) = \sqrt{3} \]
- Thus,
\[ h = \frac{\sqrt{3}}{1 - \sqrt{3}} \]

5. **Rationalize the Denominator**:
- Multiply the numerator and denominator by the conjugate of the denominator:
\[ h = \frac{\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]
- The denominator simplifies to:
\[ 1 - 3 = -2 \]
- Therefore,
\[ h = \frac{\sqrt{3} + 3}{-2} = \frac{3 + \sqrt{3}}{2} \text{ km} \]

### Final Answer:
The height of the aeroplane from the ground is:
\[ h = \frac{3 + \sqrt{3}}{2} \text{ km} \]