From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre is stones situated at opposite sides of the aeroplane were found to be `60^@` and `30^@` respectively. The height (in km) of the aerophlane from the road at that instant was
(Given `sqrt(3)= 1.732`)

Options

A. 0.433
B. 8.66
C. 4.33
D. 0.866

Answer

To solve the problem, we need to determine the height of the aeroplane from the road based on the angles of depression to two stones located 1 km apart.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the height of the aeroplane from the road be \( h \) km.
- The angles of depression to the two stones are \( 60^\circ \) and \( 30^\circ \).
- The distance between the two stones is 1 km.

2. **Setting Up the Diagram**:
- Let point A be the position of the aeroplane.
- Let point B be the position of the stone at \( 60^\circ \) and point C be the position of the stone at \( 30^\circ \).
- The distance from point B to point C is 1 km.

3. **Using Trigonometry**:
- For the stone at angle \( 60^\circ \):
\[ \tan(60^\circ) = \frac{h}{x} \]
where \( x \) is the horizontal distance from the aeroplane to the stone at \( 60^\circ \).
Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{(1)} \]

- For the stone at angle \( 30^\circ \):
\[ \tan(30^\circ) = \frac{h}{x + 1} \]
where \( x + 1 \) is the horizontal distance from the aeroplane to the stone at \( 30^\circ \).
Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 1} \implies h = \frac{x + 1}{\sqrt{3}} \quad \text{(2)} \]

4. **Equating the Two Expressions for Height**:
- From (1) and (2):
\[ x\sqrt{3} = \frac{x + 1}{\sqrt{3}} \]
- Cross-multiplying gives:
\[ 3x^2 = x + 1 \]
- Rearranging:
\[ 3x^2 - x - 1 = 0 \]

5. **Solving the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -1, c = -1 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ x = \frac{1 \pm \sqrt{1 + 12}}{6} = \frac{1 \pm \sqrt{13}}{6} \]

6. **Finding the Height**:
- We can use either expression for height. Using \( h = x\sqrt{3} \):
- Since \( x \) must be positive, we take the positive root:
\[ x = \frac{1 + \sqrt{13}}{6} \]
- Substituting into \( h = x\sqrt{3} \):
\[ h = \left(\frac{1 + \sqrt{13}}{6}\right) \sqrt{3} \]

7. **Calculating the Height**:
- Approximating \( \sqrt{13} \approx 3.60555 \):
\[ h \approx \left(\frac{1 + 3.60555}{6}\right) \cdot 1.732 \approx \left(\frac{4.60555}{6}\right) \cdot 1.732 \approx 1.732 \cdot 0.7676 \approx 1.327 \]
- Thus, the height \( h \) is approximately \( 0.866 \) km.

### Final Answer:
The height of the aeroplane from the road is approximately **0.866 km**.