7. A helicopter flies 110 km east from city A to city B, then 100 km South from city B to cityC, and finally 200 km northwest to city D. How far is it from city A to city D? In whatdirection must the airplane head to return directly to city A from city D?

To solve this problem, we can use the concept of vectors and trigonometry.

Step 1: Represent each leg of the journey as a vector.

• The first leg from city A to city B is 110 km east. We can represent this as a vector (110, 0) in a coordinate system where east is the positive x direction.
• The second leg from city B to city C is 100 km south. This can be represented as a vector (0, -100) where south is the negative y direction.
• The final leg from city C to city D is 200 km northwest. This is a bit trickier to represent as a vector because it's not along a cardinal direction. We can break it down into its north and west components. The north component is 200cos(45) = 141.42 km and the west component is 200sin(45) = 141.42 km. So the vector is (-141.42, 141.42).

Step 2: Add the vectors together to find the total displacement from city A to city D.

• The total displacement in the x direction (east-west) is 110 - 141.42 = -31.42 km. The negative sign indicates that the displacement is towards the west.
• The total displacement in the y direction (north-south) is 0 - 100 + 141.42 = 41.42 km. The positive sign indicates that the displacement is towards the north.

Step 3: Use the Pythagorean theorem to find the total distance from city A to city D.

• The total distance is sqrt((-31.42)^2 + 41.42^2) = 51.96 km.

Step 4: Use trigonometry to find the direction from city D to city A.

• The angle θ between the displacement vector and the positive x axis (east) can be found using the formula tan(θ) = y/x = 41.42/-31.42. Solving for θ gives θ = atan(41.42/-31.42) = -52.57 degrees. The negative sign indicates that the direction is west of north.

So, the helicopter is 51.96 km from city A and must head 52.57 degrees west of north to return directly to city A.