a boat leaves lighthouse A and sails 5.4km. At this time it is sighted from lighthouse B, 7.2km west of A. bearing of the boat from B is north65degrees40minuteseast. how far is the boat from B.

To solve this problem, we can use trigonometry to determine the distance of the boat from Lighthouse B. Let's go through it step-by-step.

### Step 1: Diagram and Information Summary
- Let:
  - \( A \): Position of Lighthouse A.
  - \( B \): Position of Lighthouse B.
  - \( C \): Current position of the boat.
- The information given:
  - Distance \( AB = 7.2 \) km (B is directly west of A).
  - Distance \( AC = 5.4 \) km.
  - Bearing of the boat from B is \( \text{N}65^\circ40'\text{E} \), which means the angle \( \angle ABC = 65^\circ40' \).

### Step 2: Applying the Law of Cosines
The Law of Cosines relates the sides and angles in any triangle, and in this case, we are working with triangle \( ABC \).

The formula for the Law of Cosines is:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where:
- \( c \) is the side opposite the angle \( C \),
- \( a \) and \( b \) are the other two sides.

For our triangle:
- \( a = 7.2 \) km (distance \( AB \)),
- \( b = 5.4 \) km (distance \( AC \)),
- \( C = 65^\circ40' \).

We want to find the distance \( BC \) (let's denote it as \( c \)).

### Step 3: Calculate \( BC \)
Substitute the known values into the Law of Cosines:
\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(65^\circ40')
\]

Calculating each part:
1. \( AB^2 = (7.2)^2 = 51.84 \)
2. \( AC^2 = (5.4)^2 = 29.16 \)
3. \( \cos(65^\circ40') \approx 0.4156 \) (using a calculator for cosine value in degrees)

So,
\[
BC^2 = 51.84 + 29.16 - 2 \cdot 7.2 \cdot 5.4 \cdot 0.4156
\]
\[
BC^2 = 51.84 + 29.16 - 32.34
\]
\[
BC^2 = 48.66
\]
\[
BC = \sqrt{48.66} \approx 6.98 \text{ km}
\]

### Final Answer
The boat is approximately **6.98 km** from Lighthouse B.

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Do you have any questions, or would you like further details on any part of this solution?

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Here are five related questions for further practice:
1. How would the distance from B to C change if the boat had traveled 8 km instead of 5.4 km?
2. How would you approach finding the bearing from A to the boat's current location?
3. What is the angle \( \angle BAC \) in this triangle?
4. If the bearing were N45°E instead of N65°40'E, how would the distance calculation change?
5. How would you find the shortest distance from the boat to the line joining lighthouses A and B?

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**Tip:** When solving problems involving bearings, draw a sketch to clarify directions and angles—it helps with visualizing the trigonometric setup.

A boat leaves lighthouse A and sails 5.4 km. At this time it is sighted from lighthouse B, 7.2 km west of A. The bearing of the boat from B is north 65 degrees 40 minutes east. How far is the boat from B?

Learn how to calculate the distance of a boat from Lighthouse B, given bearings and distances from two points. This problem involves trigonometric concepts and the Law of Cosines, making it suitable for students in Grades 10-12.

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