Lihaan walks 8 m towards South from his office to reach a park. He then turns…

Concept: In a direction-sense problem, track the walker's net displacement along two perpendicular axes (North-South and East-West). A left turn rotates the current facing direction 90° anticlockwise and a right turn rotates it 90° clockwise; a stated "90° anticlockwise" turn behaves exactly like a left turn. Two straight-line legs walked in exactly opposite directions cancel each other out, so only the leg(s) along the axis that does not cancel decide the final direction and distance.

Application -- tracing each leg from the office:

1. Office to park: facing South, walks 8 m South.

2. Park to traffic signal: at the park he turns left. A left turn while facing South changes the facing to East, so he walks 15 m East to the traffic signal.

3. Traffic signal to coffee shop: he turns 90° anticlockwise -- the same rotation as a left turn -- so the facing changes from East to North, and he walks 8 m North to the coffee shop.

4. Netting the two vertical (South-North) legs: the 8 m walked South (office to park) and the 8 m walked North (traffic signal to coffee shop) are equal and opposite, so they cancel completely -- the coffee shop lies on the same horizontal line as the office.

5. With the vertical component cancelled, the only surviving displacement is the 15 m Eastward leg from the park to the traffic signal, so the coffee shop is 15 m East of the office.

Cross-check: the diagram below places the office and the coffee shop on the same top edge of a rectangle, 15 m apart, with the park and the traffic signal directly below them on the bottom edge -- confirming the office and coffee shop differ only in the East-West direction, by exactly 15 m. (The final leg to home, from the coffee shop, describes a point beyond this question and does not affect the office-to-coffee-shop distance being asked.)

Therefore, the coffee shop is 15 m East of Lihaan's office.