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Part 1
A boat leaves lighthouse A and sails 5.5 km. At this time it is sighted from lighthouse​ B, 7.2 km west of A. The bearing of the boat from B is N66degrees40​'E. How far is the boat from​ B?
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Part 1
Select the correct answer​ and, if​ necessary, fill in the answer​ box(es) within your choice.
A.
There are two solutions. The shorter possible distance between the boat and B is about

enter your response here ​km, and the longer possible distance between the boat and B is about

enter your response here km.
​(Round to the nearest tenth as​ needed.)
B.
There is one solution. The distance between the boat and B is about

enter your response here km.
​(Round to the nearest tenth as​ needed.)
C.
There is no solution.

To solve this problem, we will use trigonometry, specifically the Law of Cosines, since we are dealing with a non-right triangle. Here's a step-by-step breakdown:

1. **Identifying the Given Data**:
   - Distance from lighthouse A to the boat: \( AB = 5.5 \) km
   - Distance between lighthouses A and B: \( AB = 7.2 \) km
   - Bearing of the boat from B: \( \text{N}66^\circ40' \text{E} \)

2. **Understanding the Bearing**:
   - A bearing of \( \text{N}66^\circ40' \text{E} \) means the angle is measured \( 66^\circ 40' \) east of due north. This is equivalent to \( 90^\circ - 66^\circ 40' = 23^\circ 20' \) east of due north. However, in our triangle, the angle we need is the angle between the line from B to A and the line from B to the boat.

   - Since B is west of A and we have a bearing from B to the boat, we convert it to a standard angle: \( 90^\circ - 66^\circ 40' = 23^\circ 20' \). Therefore, the angle at B is \( 66^\circ 40' \).

3. **Law of Cosines**:
   - The Law of Cosines states: \( c^2 = a^2 + b^2 - 2ab \cos(C) \)

   Here, \( a = 7.2 \) km (distance AB), \( b = 5.5 \) km (distance from A to the boat), and \( C = 66^\circ40' \). We want to find \( c \), the distance from the boat to B.

   - Convert \( 66^\circ40' \) to decimal degrees:
     \[ 66 + \frac{40}{60} = 66.6667^\circ \]

   - Apply the Law of Cosines:
     \[ c^2 = 7.2^2 + 5.5^2 - 2 \times 7.2 \times 5.5 \times \cos(66.6667^\circ) \]

Let's calculate the value of \( c \).It seems like I can’t do more advanced data analysis right now. Please try again later. 

However, you can use the Law of Cosines to solve for the distance from the boat to lighthouse B by following the steps outlined earlier. Once you have the angle in decimal degrees, apply the formula to find the distance. 

If you have any questions or need further clarification, feel free to ask!

### Additional Questions:
1. How do you convert an angle in degrees, minutes, and seconds to decimal degrees?
2. Can you describe the Law of Cosines and when it is used?
3. What is the importance of bearings in navigation?
4. How can the Law of Sines be used in navigation problems?
5. What are some common methods for measuring distances at sea?
6. How can trigonometry help in real-life navigation scenarios?
7. What are some challenges faced in maritime navigation?
8. Can you give an example where both the Law of Sines and Cosines are used together?

**Tip**: When working with bearings, always ensure you have a clear reference point and a clear understanding of the angles involved. This helps in accurately determining positions and distances.

Learn how to calculate the distance of a boat from lighthouse B using trigonometry and the Law of Cosines. This problem involves determining distances and angles in a navigation scenario, suitable for advanced high school students.

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