Overview: For both problems we'll place the starting point at the origin (0,0), track each move to get final coordinates, and then use the distance formula (which comes from the Pythagorean theorem): distance = sqrt((change in x)^2 + (change in y)^2).
Problem 1 — W's path:
- Start at (0,0).
- 60 miles north → (0,60).
- 30 miles east → (30,60).
- 30 miles north → (30,90).
- 150 miles west → (30−150, 90) = (−120, 90).
Now find the distance from the origin to (−120, 90):
distance = sqrt((−120)^2 + 90^2) = sqrt(14400 + 8100) = sqrt(22500) = 150.
Answer 1: W is 150 miles from the starting point.
Problem 2 — positions of A, B, C, D, E, F, G (in meters):
Choose B at (0,0) to make calculations easy.
- A is 50 m east of B → A is at x = 0 + 50 = 50.
- A is 30 m west of C → C is at x = A_x + 30 = 50 + 30 = 80.
- D is 60 m east of C → D_x = 80 + 60 = 140.
- D is 40 m east of E → E_x = D_x − 40 = 140 − 40 = 100.
So E is at (100, ?). Since no north/south position was given for E, set E_y = 0 for convenience. Then:
- F is 50 m north of E → F is at (100, 50).
- F is 80 m north of G → G is 80 m south of F, so G is at (100, 50 − 80) = (100, −30).
B is at (0,0) and G is at (100, −30). Use the distance formula:
distance BG = sqrt((100 − 0)^2 + (−30 − 0)^2) = sqrt(100^2 + (−30)^2) = sqrt(10000 + 900) = sqrt(10900) ≈ 104.403... m.
Answer 2: To the nearest tenth of a meter, B and G are 104.4 m apart.
Tip: Placing one known point at (0,0) and tracking coordinates step-by-step makes these problems much easier. The Pythagorean theorem then gives the straight-line distance between any two points.