91Ó°ÊÓ

To solve this problem, we can represent Orville's movements as vectors. Let A be his starting point, B be the point where he changes direction, and C be his final destination (200 meters northeast of point A). We can represent the first movement as the vector AB and the second movement as the vector BC. First movement (AB): 320 m due east Second movement (BC): unknown distance and direction

Set up a coordinate system, with the origin at point A, the x-axis pointing east, and the y-axis pointing north. In this coordinate system, the coordinates of points A, B, and C can be written as: A = (0,0) B = (320,0) C = (x, y) Now, we know that point C is 200 meters northeast of point A; thus, the displacement AC can be represented as: AC = (200 cos(45°), 200 sin(45°)) = (100√2, 100√2)

Using vector addition, we can relate the vectors AC, AB, and BC: AC = AB + BC (100√2, 100√2) = (320, 0) + BC We can now find the coordinates of point C by solving for BC: BC = (100√2, 100√2) - (320, 0) BC = (100√2 - 320, 100√2)

To find the distance and direction of the second movement (BC), we need to calculate the magnitude and angle of vector BC. The magnitude can be calculated using the Pythagorean theorem: \(Distance = |BC| = \sqrt{(100\sqrt{2} - 320)^2 + (100\sqrt{2})^2}\) \(Distance ≈ 324.50 m\) The direction can be found using the arctangent function: \(Direction = tan^{-1}\left(\frac{y}{x}\right) = tan^{-1}\left(\frac{100\sqrt{2}}{100\sqrt{2} - 320}\right)\) \(Direction ≈ 53.31°\) Since we are working in a coordinate system where the x-axis points east and the y-axis points north, this angle represents the direction measured counterclockwise from east. So, the second portion of the trip is approximately 324.50 meters long and in a direction approximately 53.31° counterclockwise from east.