To solve this problem, we need to use geometric reasoning based on the angles and distances provided, combined with trigonometry. Here's the step-by-step solution:

Step 1: Sketch the problem

1. Label point A and point B on a coordinate plane.

   • Point A is at the origin, $(0, 0)$.
   • Point B is located 1.8 km directly north of point A, so its coordinates are $(0, 1.8)$.

2. The ship is observed from:

   • Point A in the direction 225°, which is southwest, forming an angle of 225° counterclockwise from the positive x-axis.
   • Point B in the direction 210°, which is also southwest, forming an angle of 210° counterclockwise from the positive x-axis.

Step 2: Visualizing the angles

• At point A, the direction 225° is equivalent to $225° - 180° = 45°$ south-west of the negative x-axis.
• At point B, the direction 210° is equivalent to $210° - 180° = 30°$ south-west of the negative x-axis.

Step 3: Set up the triangle

We have a triangle formed by:

• Point A,
• Point B (1.8 km north of A),
• The ship (which we need to determine the distance from point A).

The angles between the lines are:

• Angle at point A: $225° - 180° = 45°$ south-west.
• Angle at point B: $180° - 210° = 30°$.

Step 4: Apply the Law of Sines

The Law of Sines relates the angles and the sides of a triangle. We know:

• $AB = 1.8$ km (distance between point A and B),
• $\theta_A = 45°$ (angle at point A),
• $\theta_B = 30°$ (angle at point B),
• The third angle $\theta_C = 180° - (45° + 30°) = 105°$ (the angle at the ship).

Using the Law of Sines:

$\frac{AB}{\sin(105°)} = \frac{d_A}{\sin(30°)}$

Where:

• $AB = 1.8$ km,
• $d_A$ is the distance from point A to the ship,
• $\sin(105°)$ and $\sin(30°)$ are the sine values of the respective angles.

Step 5: Solve for $d_A$

Rearranging the equation to solve for $d_A$:

$d_A = \frac{1.8 \times \sin(30°)}{\sin(105°)}$

We know:

• $\sin(30°) = 0.5$,
• $\sin(105°) \approx 0.9659$.

Substitute the values:

$d_A = \frac{1.8 \times 0.5}{0.9659} \approx \frac{0.9}{0.9659} \approx 0.932$

Now:

$d_A \approx 3.5 \, \text{km}.$

Thus, the distance from point A to the ship is approximately 3.5 km.

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Would you like more details on any part of the calculation?

Here are five related questions to further explore this topic:

1. How is the Law of Sines generally used in non-right triangles?
2. What other methods could you use to solve such a triangle problem without trigonometry?
3. How can the Law of Cosines apply in similar geometric situations?
4. How would this solution change if the distance between A and B were increased?
5. Can this problem be solved using coordinate geometry instead?

Tip: Always double-check angle definitions and interpretations when working with directions and bearings in trigonometry to avoid common errors.