The angles of elevation of an artificial satellite measured from two earth stations are `30^@ and 60^@` respectively. If the distance between the earth stations, which are in straight line with the point directly below the satellite, is 4000 km, then the height of the satellite is ______ - (Use `sqrt3 = 1.732`)

Options

A. 2000 km
B. 6000 km
C. 3464 km
D. 2828 km

Answer

To find the height of the satellite, we will use the information given about the angles of elevation and the distance between the two earth stations.

### Step-by-Step Solution:

1. **Identify the problem setup**:
- Let the height of the satellite be \( h \).
- Let the distance between the two earth stations (point C and D) be \( 4000 \) km.
- The angles of elevation from the two stations are \( 30^\circ \) and \( 60^\circ \).

2. **Draw the diagram**:
- Draw a vertical line representing the height \( h \) of the satellite.
- Label the point directly below the satellite on the ground as point B.
- Label the two earth stations as point C (where the angle of elevation is \( 60^\circ \)) and point D (where the angle of elevation is \( 30^\circ \)).
- The distance between points C and D is \( CD = 4000 \) km.

3. **Set up the triangles**:
- Triangle \( ABD \) (where angle \( DAB = 30^\circ \)):
- From the definition of tangent, we have:
\[ \tan(30^\circ) = \frac{AB}{DB} \]
Here, \( AB = h \) and \( DB = 4000 + x \) (where \( x \) is the distance from point B to point C).

- Triangle \( ABC \) (where angle \( CAB = 60^\circ \)):
- Similarly, we have:
\[ \tan(60^\circ) = \frac{AB}{CB} \]
Here, \( CB = x \).

4. **Calculate the tangents**:
- We know:
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(60^\circ) = \sqrt{3} \]
- For triangle \( ABD \):
\[ \frac{1}{\sqrt{3}} = \frac{h}{4000 + x} \implies h = \frac{4000 + x}{\sqrt{3}} \]
- For triangle \( ABC \):
\[ \sqrt{3} = \frac{h}{x} \implies h = x \sqrt{3} \]

5. **Equate the two expressions for \( h \)**:
- Set the two expressions for \( h \) equal to each other:
\[ \frac{4000 + x}{\sqrt{3}} = x \sqrt{3} \]

6. **Clear the fraction**:
- Multiply both sides by \( \sqrt{3} \):
\[ 4000 + x = 3x \]
- Rearranging gives:
\[ 4000 = 3x - x \implies 4000 = 2x \implies x = 2000 \]

7. **Substitute back to find \( h \)**:
- Substitute \( x = 2000 \) back into the equation for \( h \):
\[ h = 2000 \sqrt{3} \]
- Using \( \sqrt{3} \approx 1.732 \):
\[ h = 2000 \times 1.732 = 3464 \text{ km} \]

### Final Answer:
The height of the satellite is approximately \( 3464 \) km.