To solve this problem, we need to calculate the distance between the airplane and the airport after completing the round trip with the navigation error.

Step 1: Represent the problem

1. The plane first travels 100 miles on a course of $30^\circ$ east of south. This means the direction vector is: $(x_1, y_1) = (100 \sin(30^\circ), -100 \cos(30^\circ)) = (50, -86.6)$

2. The plane then flies 100 miles on a course of $40^\circ$ west of north. This direction vector is: $(x_2, y_2) = (-100 \sin(40^\circ), 100 \cos(40^\circ)) = (-64.28, 76.60)$

Step 2: Add the displacements

The total displacement vector is: $(x_{\text{total}}, y_{\text{total}}) = (x_1 + x_2, y_1 + y_2)$ Substituting values: $x_{\text{total}} = 50 - 64.28 = -14.28 \quad \text{and} \quad y_{\text{total}} = -86.6 + 76.6 = -10$

Step 3: Calculate the resultant distance

The distance of the plane from the airport is the magnitude of the displacement vector: $\text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}$ $\text{Distance} = \sqrt{(-14.28)^2 + (-10)^2} = \sqrt{203.93 + 100} = \sqrt{303.93} \approx 17.44 \, \text{miles}$

Final Answer:

The airplane is approximately 17.44 miles away from the airport.

Let me know if you'd like further clarification or additional steps! Here are five related questions for exploration:

1. What if the second leg had a different angle of flight?
2. How does the calculation change if the distances traveled are different?
3. Can you determine the airplane's exact heading relative to the airport?
4. How does vector addition simplify solving navigation problems like this?
5. What is the role of trigonometry in solving navigation and displacement problems?

Tip: Always sketch a diagram for such problems to better visualize the direction and magnitude of the vectors.