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Difficulty: HardDirection and Distance Test

At sunrise, an agricultural inspector standing at the entrance gate of a research facility observes that his shadow is cast directly behind him. Maintaining this initial facing direction, he walks 10 m10\text{ m} straight ahead. He then turns 135∘135^\circ clockwise and walks 82 m8\sqrt{2}\text{ m}. Next, he turns 135∘135^\circ anti-clockwise and walks 10 m10\text{ m}. Finally, he turns left and walks 3 m3\text{ m} to reach the observation station. What is the shortest distance between his starting position and the observation station, and in which direction is the observation station relative to his starting position?

  1. 13 m13\text{ m}, South-EastAnswer
  2. B
    13 m13\text{ m}, North-West
  3. C
    13 m13\text{ m}, North-East
  4. D
    17 m17\text{ m}, South-East

Answer

The shortest distance between the starting position and the observation station is 13 m13\text{ m}, and its direction relative to the starting point is South-East.
At sunrise, the sun rises in the East, so a shadow cast directly behind a person indicates that the person is facing East. Starting from the origin (0,0)(0,0) facing East: walking 10 m10\text{ m} East reaches (10,0)(10,0). Turning 135∘135^\circ clockwise redirects the path towards South-West; moving 82 m8\sqrt{2}\text{ m} in this direction subtracts 8 m8\text{ m} from both xx and yy coordinates, reaching (2,−8)(2,-8). Turning 135∘135^\circ anti-clockwise restores facing East; walking 10 m10\text{ m} East moves the location to (12,−8)(12,-8). Finally, turning left faces North, and walking 3 m3\text{ m} North brings the final position to (12,−5)(12,-5). Applying the Pythagorean theorem yields 122+(−5)2=13 m\sqrt{12^2 + (-5)^2} = 13\text{ m}. Because the final position has a positive East component (+12+12) and a negative North component (−5-5, which is South), the direction relative to the start is South-East.

Step-by-Step Solution

1
Determine initial facing direction from shadow scenario
At sunrise, the Sun is in the East. If a person's shadow is cast directly behind him, he must be facing East.
Light rays coming from the East cast shadows pointing West (opposite direction).
2
Trace movements using Cartesian coordinates starting from origin O(0,0)O(0, 0)
Step 1: Walk 10 m10\text{ m} East →P1=(10,0)\rightarrow P_1 = (10, 0). Facing East (0∘0^\circ).
Step 2: Turn 135∘135^\circ clockwise →\rightarrow Facing South-West (225∘225^\circ). Walk 82 m8\sqrt{2}\text{ m} SW:
Δx=−82cos⁡(45∘)=−8\Delta x = -8\sqrt{2} \cos(45^\circ) = -8, Δy=−82sin⁡(45∘)=−8\Delta y = -8\sqrt{2} \sin(45^\circ) = -8.
New position P2=(10−8,0−8)=(2,−8)P_2 = (10 - 8, 0 - 8) = (2, -8).
Step 3: Turn 135∘135^\circ anti-clockwise from SW →\rightarrow Facing East (0∘0^\circ). Walk 10 m10\text{ m} East:
New position P3=(2+10,−8)=(12,−8)P_3 = (2 + 10, -8) = (12, -8).
Step 4: Turn left from East →\rightarrow Facing North (90∘90^\circ). Walk 3 m3\text{ m} North:
Final position P4=(12,−8+3)=(12,−5)P_4 = (12, -8 + 3) = (12, -5).
Breaking down movements into orthogonal vector components (xx for East-West, yy for North-South) allows exact calculation of net displacement.
3
Calculate shortest straight-line distance and relative direction
Net coordinates = (12,−5)(12, -5).
Shortest distance d=122+(−5)2=144+25=169=13 md = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}.
Since x>0x > 0 (East) and y<0y < 0 (South), the final location is South-East of the starting point.
The Pythagorean theorem gives the displacement magnitude, and coordinate signs determine the directional quadrant.

Key Concept

Direction and Distance Test with Angular Rotations and Implicit Shadow Directions
Estimated Time:2m 30s
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Topics

Direction and Distance Test

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