This problem involves a triangle formed by the dock and the two hotels. The three sides of the triangle are the distances between the dock and the two hotels, and the angle between the lines connecting the dock to each hotel. Since we know the angles and the distance between the hotels, we can use the Law of Sines to solve for the distances.

Problem Summary:

• The distance between the two hotels: $c = 550 \, \text{m}$
• Angle at the dock between Hotel 1 and Hotel 2: $\angle A = 49^\circ$
• Angle at the dock between Hotel 2 and Hotel 1: $\angle B = 56^\circ$

We need to find the distances from the dock to each hotel:

• Distance to Hotel 1: $a$
• Distance to Hotel 2: $b$

First, let's calculate the remaining angle at the dock: $\angle C = 180^\circ - 49^\circ - 56^\circ = 75^\circ$

Now we can use the Law of Sines: $\frac{a}{\sin B} = \frac{b}{\sin A} = \frac{c}{\sin C}$

Using this, we can solve for $a$ and $b$: $a = \frac{c \cdot \sin B}{\sin C}, \quad b = \frac{c \cdot \sin A}{\sin C}$

Let's calculate the distances now.The distances from the dock to each hotel are approximately:

• Distance to Hotel 1: $a = 472.06 \, \text{m}$
• Distance to Hotel 2: $b = 429.73 \, \text{m}$

Would you like more details or have any questions? Here are some related questions:

1. How does the Law of Sines work in other triangle problems?
2. What if the distance between the hotels was different? How would it affect the solution?
3. Can we solve this problem using the Law of Cosines instead? How would the approach change?
4. What is the importance of angle measures in real-world navigation problems like this?
5. How would the solution change if the angles were given in radians?

Tip: Always verify the sum of angles in a triangle equals $180^\circ$ to avoid calculation errors.