The angles of elevation of an aeroplane flying vertically above the ground,as observed from the two consecutive stones,I km apart, are `45^(@)` and`60^(@)`aeroplane from the ground .The height of the aeroplane above the ground in km 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 solve the problem, we need to find the height of the aeroplane above the ground based on the angles of elevation observed from two points that are 1 km apart. Let's denote the height of the aeroplane as \( h \) km.

### Step-by-Step Solution:

1. **Understanding the Setup:**
- Let point A be the position of the first observer (stone) and point B be the position of the second observer, which is 1 km away from point A.
- The angle of elevation from point A is \( 45^\circ \) and from point B is \( 60^\circ \).

2. **Using Trigonometric Ratios:**
- From point A (angle \( 45^\circ \)):
\[ \tan(45^\circ) = \frac{h}{x} \quad \text{(where \( x \) is the horizontal distance from A to the point directly below the aeroplane)} \]
Since \( \tan(45^\circ) = 1 \):
\[ h = x \quad \text{(1)} \]

- From point B (angle \( 60^\circ \)):
\[ \tan(60^\circ) = \frac{h}{x + 1} \quad \text{(since B is 1 km further from A)} \]
Since \( \tan(60^\circ) = \sqrt{3} \):
\[ h = \sqrt{3}(x + 1) \quad \text{(2)} \]

3. **Equating the Two Expressions for \( h \):**
- From equations (1) and (2), we have:
\[ x = \sqrt{3}(x + 1) \]

4. **Solving for \( x \):**
- Expanding the equation:
\[ x = \sqrt{3}x + \sqrt{3} \]
- Rearranging gives:
\[ x - \sqrt{3}x = \sqrt{3} \]
\[ x(1 - \sqrt{3}) = \sqrt{3} \]
- Thus:
\[ x = \frac{\sqrt{3}}{1 - \sqrt{3}} \quad \text{(3)} \]

5. **Substituting \( x \) Back to Find \( h \):**
- Substitute \( x \) from equation (3) into equation (1):
\[ h = \frac{\sqrt{3}}{1 - \sqrt{3}} \]

6. **Rationalizing the Denominator:**
- Multiply numerator and denominator by the conjugate:
\[ h = \frac{\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{\sqrt{3} + 3}{1 - 3} = \frac{\sqrt{3} + 3}{-2} \]
- Thus:
\[ h = \frac{3 + \sqrt{3}}{2} \]

7. **Final Answer:**
- The height of the aeroplane above the ground is:
\[ h = \frac{3 + \sqrt{3}}{2} \text{ km} \]