A boy is riding his motorcycle on a road that runs east and west. He leaves the road at a service station and rides 5.50 miles in the direction \( \mathrm{N} 16.5^{\circ} \mathrm{E} \). Then he turns to his right and rides 7.25 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.
miles

Step 1 ：Let A be the starting point, B be the point where the boy turns right. Then form a triangle \(\triangle ABP\) , where P is where the motorcycle breaks down.

Step 2 ：Use the law of cosines: \(AP^2 = AB^2 + BP^2 - 2(AB)(BP)cos(\theta)\) , where \(\theta\) is the angle at B between the path of AB and BP. However, we don't know \(\theta\) directly, we can find \(\alpha = 180^{\circ} - \theta\), where \(\alpha\) is the angle between AB and the road.

Step 3 ：Using \(\alpha = 180^{\circ} - 16.5^{\circ}\), we have \(\alpha = 163.5^{\circ}\). Thus, \(\theta = 16.5^{\circ}\).

Step 4 ：Now use the law of cosines with sides AB = 5.50 miles, BP = 7.25 miles, and \(\theta = 16.5^{\circ}\) to find AP.

Step 5 ：\(AP^2 = (5.50)^2 + (7.25)^2 - 2(5.50)(7.25)cos(16.5^{\circ})\)

Step 6 ：By calculating, we have \(AP \approx 7.31\) miles.