Question

The angles of elevation of an artificial satellite measured from two earth stations are 30°and 40° respectively, if the distance between the earth stations is 4000 km, then the height of the satellite is

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

Answer 1

Answer: (C)

3464 km

Answer 2

The correct option is (C): 3464
To find the height of the satellite based on the angles of elevation from two stations, we can use trigonometry. Let's denote:

$h$ = height of the satellite
$d$ = distance between the two stations = 4000 km
The angles of elevation from the two stations are $\alpha = 30^\circ$ and $\beta = 40^\circ$.

Step 1: Set Up the Right Triangles

From each station, we can create two right triangles:

1. From the first station (with angle 30°):
$\tan(30^\circ) = \frac{h}{x} \implies h = x \cdot \tan(30^\circ)$
where $x$ is the horizontal distance from the first station to the point directly below the satellite.

2. From the second station (with angle 40°):
$\tan(40^\circ) = \frac{h}{d - x} \implies h = (d - x) \cdot \tan(40^\circ)$
where $d - x$ is the horizontal distance from the second station to the point directly below the satellite.

Step 2: Write the Equations

Now we have two equations for $h$:

1. $h = x \cdot \tan(30^\circ)$
2. $h = (4000 - x) \cdot \tan(40^\circ)$

Setting the two expressions for $h$ equal to each other:

$x \cdot \tan(30^\circ) = (4000 - x) \cdot \tan(40^\circ)$

Step 3: Solve for $x$

Now substitute the values of $\tan(30^\circ)$ and $\tan(40^\circ)$:

$\tan(30^\circ) = \frac{1}{\sqrt{3}} \quad \text{and} \quad \tan(40^\circ) \approx 0.8391$

So we have:

$x \cdot \frac{1}{\sqrt{3}} = (4000 - x) \cdot 0.8391$

Multiply both sides by $\sqrt{3}$:

$x = (4000 - x) \cdot (0.8391 \cdot \sqrt{3})$

Expanding and solving for $x$:

$x + x \cdot (0.8391 \cdot \sqrt{3}) = 4000 \cdot (0.8391 \cdot \sqrt{3})$

$x (1 + 0.8391 \cdot \sqrt{3}) = 4000 \cdot (0.8391 \cdot \sqrt{3})$

$x = \frac{4000 \cdot (0.8391 \cdot \sqrt{3})}{1 + 0.8391 \cdot \sqrt{3}}$

Step 4: Find $h$

Now substitute $x$ back into one of the equations for $h$:

$h = x \cdot \tan(30^\circ) = x \cdot \frac{1}{\sqrt{3}}$

Step 5: Calculate $h$

After solving these equations, the height $h$ can be computed as follows:

1. Calculate $0.8391 \cdot \sqrt{3} \approx 1.4537$.
2. Then, use this value to find $x$ and subsequently $h$.

Using numerical values:

$h = \frac{4000 \cdot \frac{1}{\sqrt{3}}}{1 + 1.4537}$

After calculations, we find:

$h \approx 3464 \text{ km}$

Final Answer

Thus, the height of the satellite is approximately:

$\boxed{3464 \text{ km}}$