A boy is riding his motorcycle on a road that runs east and west. He leaves the road at a service station and rides
6.50
miles in the direction N
16.5
°
E. Then he turns to his right and rides
7.50
miles back to the road, where his motorcycle breaks down. How far will he have to walk to get back to the service station? Round your answer to 2 decimal places. Note: none of the angles between his paths are necessarily right.

To solve this problem, we can use the Law of Cosines, since none of the angles between the paths are necessarily right angles.

### Step 1: Visualize the problem
The boy's journey can be described as a triangle:
1. He first travels 6.50 miles in the direction N 16.5° E.
2. Then, he turns to his right and travels 7.50 miles back to the road.

We need to find the distance he needs to walk back to the service station.

### Step 2: Determine the angles
Given:
- He first travels in the direction N 16.5° E.
- Then, he turns to his right. Since he turns right, the angle between his initial direction and his final direction is \(90^\circ - 16.5^\circ = 73.5^\circ\).

So, the angle between the two paths (at the vertex of the triangle formed by his journey) is \(180^\circ - 73.5^\circ = 106.5^\circ\).

### Step 3: Use the Law of Cosines
The Law of Cosines formula is:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
Where:
- \(c\) is the distance back to the service station (what we want to find).
- \(a = 6.50\) miles (the first leg of his journey).
- \(b = 7.50\) miles (the second leg of his journey).
- \(C = 106.5^\circ\) (the angle between his paths).

Now, substitute the known values into the formula:
\[
c^2 = (6.50)^2 + (7.50)^2 - 2(6.50)(7.50) \cdot \cos(106.5^\circ)
\]

### Step 4: Calculate the distance
Let's compute this.

\[
c^2 = 42.25 + 56.25 - 97.5 \cdot \cos(106.5^\circ)
\]
\[
c^2 = 98.5 - 97.5 \cdot \cos(106.5^\circ)
\]
\[
c = \sqrt{98.5 - 97.5 \cdot (-0.277)} 
\]
\[
c = \sqrt{98.5 + 27.03} 
\]
\[
c = \sqrt{125.53} 
\]
\[
c \approx 11.21 \text{ miles}
\]

So, the distance he will have to walk to get back to the service station is approximately **11.21 miles**.

Would you like more details on this solution, or do you have any questions?

### Related Questions:
1. How do you apply the Law of Cosines in various types of triangles?
2. What other methods could be used to solve this problem if it were a right triangle?
3. How would the solution change if the angle between the paths were different?
4. What is the importance of knowing the direction angles in navigation problems?
5. How does the Law of Sines relate to the Law of Cosines in solving triangle problems?

### Tip:
When dealing with navigation or directional problems, always draw a diagram to help visualize the paths and angles involved. This makes it easier to apply trigonometric rules correctly.

Learn how to calculate the distance a boy needs to walk back to a service station after riding his motorcycle using the Law of Cosines. Detailed step-by-step solution with trigonometric applications. Round your answer to 2 decimal places.

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