The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 55 mi west to another airport to find that conditions there are too icy for him to land. He flies 25 mi at \(15^{\circ}\) east of south and is finally able to land at the third airport. (a) How far and in what direction must he fly the next day to go directly to his original destination? (b) How many extra miles beyond his original flight plan has he flown?

To find the net displacement, we can break down the three flights into components and sum up the components to find the net displacement. 1. Flight 1: The pilot flies 55 mi west Horizontal displacement (west) = -55 mi (assuming east as positive and west as negative) 2. Flight 2: The pilot encounters ice and cannot land, so he flies 25 mi at \(15^{\circ}\) east of south. Horizontal displacement (east) = 25 mi * cos(\(15^{\circ}\)) = 25 mi * 0.9659 = 24.15 mi Vertical displacement (south) = 25 mi * sin(\(15^{\circ}\)) = 25 mi * 0.2588 = 6.47 mi (assuming north as positive and south as negative) Now, we will add up the displacements. Total horizontal displacement = -55 + 24.15 = -30.85 mi (west) Total vertical displacement = -6.47 mi (south) We will use the Pythagorean theorem to find the net displacement (distance). Distance = \(\sqrt{(-30.85)^2 + (-6.47)^2}\) = 31.7 mi (rounded to one decimal place) Finally, we will find the direction using the tangent function. Direction = arctan(\(\frac{-6.47}{-30.85}\)) = 11.9\(^{\circ}\) (rounded to one decimal place) So the pilot must fly 31.7 miles at 11.9\(^{\circ}\) west of south to reach his original destination.

To find the extra miles the pilot has flown, we will simply add the three distances he has traveled. Extra miles = 55 mi (west) + 25 mi (east of south) + 31.7 mi (west of south) = 111.7 miles Therefore, the pilot has flown 111.7 miles extra beyond his original flight plan.