Question 1168441
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.