SOLUTION: Two long-range navigation stations A and B lie on a line running east and west, and A is 88 miles due east of B. An airplane is travelling east on a straight line course that i

Question 1169787: Two long-range navigation stations A and B lie on a line running east and west,
and A is 88 miles due east of B. An airplane is travelling east on a straight line course
that is 66 miles north of the line tough A and B. Signals are sent at the same time
from A and B, and the signal from A reaches the plane 350 microseconds before
the one from B. If the signals travel at the rate of 0.2 mile/microsecond, locate the
position of the plane.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this navigation problem step-by-step.
**Understanding the Problem**
We have two navigation stations A and B, and an airplane flying on a parallel course. We're given the distance between A and B, the airplane's distance from the line AB, the time difference in signal reception, and the signal speed. We need to find the airplane's position.
**Diagram**
1. Draw a horizontal line representing the line passing through stations A and B.
2. Mark point B on the left and point A on the right, with a distance of 88 miles between them.
3. Draw a horizontal line 66 miles above the line AB, representing the airplane's path.
4. Let P be the position of the airplane.
5. Draw lines PA and PB.
6. Draw a line from P perpendicular to AB, intersecting AB at point C.
**Given Information**
* Distance AB = 88 miles
* Airplane's distance from line AB (PC) = 66 miles
* Time difference (Δt) = 350 microseconds
* Signal speed (v) = 0.2 mile/microsecond
**Solution**
1. **Distance Difference:**
* The difference in distances traveled by the signals is:
* Δd = v * Δt = 0.2 miles/microsecond * 350 microseconds = 70 miles
* Therefore, PB - PA = 70 miles.
2. **Coordinates:**
* Let B be the origin (0, 0).
* Then A is at (88, 0).
* Let P be at (x, 66).
* Then C is at (x, 0).
3. **Distances PA and PB:**
* PA = √((x - 88)² + 66²)
* PB = √(x² + 66²)
4. **Equation:**
* PB - PA = 70
* √(x² + 66²) - √((x - 88)² + 66²) = 70
5. **Solve for x:**
* √(x² + 66²) = 70 + √((x - 88)² + 66²)
* Square both sides:
* x² + 66² = 4900 + 140√((x - 88)² + 66²) + (x - 88)² + 66²
* x² = 4900 + 140√((x - 88)² + 66²) + x² - 176x + 88²
* 176x - 4900 - 88² = 140√((x - 88)² + 66²)
* 176x - 4900 - 7744 = 140√((x - 88)² + 4356)
* 176x - 12644 = 140√((x - 88)² + 4356)
* (176x - 12644) / 140 = √((x - 88)² + 4356)
* (44x - 3161) / 35 = √((x - 88)² + 4356)
* Square both sides again:
* (44x - 3161)² / 35² = (x - 88)² + 4356
* (1936x² - 278128x + 9992081) / 1225 = x² - 176x + 7744 + 4356
* (1936x² - 278128x + 9992081) / 1225 = x² - 176x + 12100
* 1936x² - 278128x + 9992081 = 1225(x² - 176x + 12100)
* 1936x² - 278128x + 9992081 = 1225x² - 215600x + 14822500
* 711x² - 62528x - 4830419 = 0
* Solve the quadratic equation using the quadratic formula:
* x = (62528 ± √(62528² - 4 * 711 * (-4830419))) / (2 * 711)
* x = (62528 ± √(3909775384 + 13739773956)) / 1422
* x = (62528 ± √17649549340) / 1422
* x = (62528 ± 132851.61) / 1422
* x ≈ 136.69 or x ≈ -48.68
* Since the plane is traveling east, we choose the positive solution.
* x ≈ 136.69
6. **Position of the Plane:**
* The plane's position is approximately (136.69, 66).
* This means the plane is approximately 136.69 miles east of station B.
**Final Answer**
The plane is approximately 136.69 miles east of station B.

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