The path of a satellite orbiting the earth causes the satellite to pass directly over two tracking stations A and B, which are 60 mi apart. When the satellite is on one side of the two stations, the angles of elevation at A and B are measured to be 87.0° and 84.2°, respectively. (Round your answers to the nearest mile.)

To solve the problem, we can model the situation using geometry and trigonometry. Here, the satellite's position forms a triangle with the tracking stations A and B on the ground. We can use the law of sines to calculate the altitude of the satellite.

### Step-by-step solution:

1. **Define the variables:**
   - Let \( h \) be the altitude of the satellite.
   - Let \( d_A \) and \( d_B \) be the distances from the satellite to stations A and B, respectively.
   - The distance between A and B is given as 60 miles.
   - The angles of elevation at A and B are 87.0° and 84.2°, respectively.

2. **Label the triangle:**
   The triangle formed by the satellite and the two stations is a non-right triangle:
   - At station A, the angle of elevation is 87.0°, so the angle at the ground level adjacent to A (i.e., the angle opposite \( d_B \)) is \( 180^\circ - 87^\circ = 93^\circ \).
   - At station B, the angle of elevation is 84.2°, so the angle at the ground level adjacent to B (i.e., the angle opposite \( d_A \)) is \( 180^\circ - 84.2^\circ = 95.8^\circ \).

3. **Calculate the remaining angle in the triangle:**
   The angle at the satellite is the remaining angle in the triangle, which can be found as:
   \[
   \text{Angle at satellite} = 180^\circ - 93^\circ - 95.8^\circ = 91.2^\circ
   \]

4. **Use the Law of Sines:**
   We can apply the law of sines to find the distances \( d_A \) and \( d_B \):
   \[
   \frac{60}{\sin(91.2^\circ)} = \frac{d_A}{\sin(95.8^\circ)} = \frac{d_B}{\sin(93^\circ)}
   \]

   Let's calculate \( d_A \) and \( d_B \) first.

5. **Calculate the altitude:**
   Once we have \( d_A \) and \( d_B \), we can use the trigonometric relation \( h = d_A \sin(87^\circ) \) or \( h = d_B \sin(84.2^\circ) \) to find the altitude of the satellite.

I'll now compute these values.It appears there was an issue with the calculation, resulting in negative values for the distances and altitude. I'll correct the calculations and try again.It seems that the negative values persist due to an error in interpreting the angles of elevation. I'll review the setup once more to ensure accuracy before recalculating. The angles provided are very close to vertical (87.0° and 84.2°), meaning the distances and the geometry involved need to be handled carefully. Let me try an alternate approach.Let's take a step back and carefully reanalyze the problem geometry.

The negative values suggest that something is off in the configuration of the angles. Since angles close to 90° (like 87° and 84.2°) indicate a nearly vertical satellite position, we need to ensure the correct angle handling in our trigonometric calculations.

I will take a more refined approach to re-calculate the satellite's altitude.The altitude of the satellite is approximately **1,220 miles** above the Earth.

Would you like further details or have any questions? Here are five related questions to explore:

1. How would the altitude change if the distance between tracking stations A and B increased?
2. What if the angles of elevation were closer to 45° instead of near-vertical?
3. How can the law of sines be applied to other satellite tracking problems?
4. What is the significance of using right triangles in this type of problem?
5. How does the satellite's speed affect its orbital altitude?

**Tip:** In satellite orbit problems, angles of elevation close to 90° indicate that the satellite is almost directly overhead, simplifying altitude calculations.

Solve a satellite tracking problem involving two stations 60 miles apart with angles of elevation of 87.0° and 84.2°. Learn how to apply the law of sines and trigonometric functions to calculate the satellite's altitude. Suitable for students in Grades 10-12.

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