SOLUTION: Two Coast Guard stations, A and B, are on an east-west line and are 115 km apart. The bearing of a ship from station A is N 18o E, and the bearing of the same ship from station B

Question 1168441: Two Coast Guard stations, A and B, are on an east-west line and are 115 km apart. The bearing of a ship from station A is N 18o E, and the bearing of the same ship from station B is N 33o W.
a) How far is the ship from the east-west line connecting the two Coast Guard stations.
b) How far is the ship from station A?
c) How far is the ship from station B?

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**1. Visualize the Situation**
* Draw a horizontal line representing the east-west line connecting stations A and B.
* Place A and B on this line, 115 km apart, with A to the west and B to the east.
* Draw a line from A in the direction N 18° E.
* Draw a line from B in the direction N 33° W.
* The intersection of these two lines is the position of the ship (S).
**2. Set Up the Triangle**
* Let S be the position of the ship.
* We have triangle ABS.
* AB = 115 km
* Angle SAB = 90° - 18° = 72°
* Angle SBA = 90° - 33° = 57°
* Angle ASB = 180° - (72° + 57°) = 180° - 129° = 51°
**3. Use the Law of Sines**
We can use the Law of Sines to find the distances AS and BS:
* AS / sin(57°) = BS / sin(72°) = AB / sin(51°)
**b) How far is the ship from station A? (AS)**
* AS / sin(57°) = 115 / sin(51°)
* AS = 115 * sin(57°) / sin(51°)
* AS ≈ 115 * 0.8387 / 0.7771
* AS ≈ 124.06 km
**c) How far is the ship from station B? (BS)**
* BS / sin(72°) = 115 / sin(51°)
* BS = 115 * sin(72°) / sin(51°)
* BS ≈ 115 * 0.9511 / 0.7771
* BS ≈ 140.73 km
**a) How far is the ship from the east-west line connecting the two Coast Guard stations?**
Let's call the point where the perpendicular from S to AB meets AB as point P. We need to find the length of SP.
We can use triangle ASP to find SP.
* sin(72°) = SP / AS
* SP = AS * sin(72°)
* SP ≈ 124.06 * sin(72°)
* SP ≈ 124.06 * 0.9511
* SP ≈ 118 km
Alternatively, we can use triangle BSP.
* sin(57°) = SP / BS
* SP = BS * sin(57°)
* SP ≈ 140.73 * sin(57°)
* SP ≈ 140.73 * 0.8387
* SP ≈ 118 km
**Answers:**
a) The ship is approximately 118 km from the east-west line connecting the two Coast Guard stations.
b) The ship is approximately 124.06 km from station A.
c) The ship is approximately 140.73 km from station B.

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