A man is initially facing towards the east. He turns towards the north and walks 5 km, and then turns 270 degrees in the anticlockwise direction and walks another 12 km. What is the minimum straight line distance between his initial starting position and his final position?

Introduction / Context:
This question combines angle based turning with distance movement, and it asks for the minimum distance between the initial and final positions. This minimum distance is the straight line distance between the two locations, which can be found using basic geometry once we know the coordinates of the final point. Understanding how a 270 degree anticlockwise turn affects direction is crucial here.

Given Data / Assumptions:

• The man is initially facing east.
• He turns towards the north and walks 5 km.
• After that he turns 270 degrees anticlockwise.
• He then walks another 12 km in the new direction after the 270 degree anticlockwise turn.
• The ground is taken as flat and all moves are straight line segments.

Concept / Approach:
First we determine his direction after each turn. A turn of 90 degrees anticlockwise changes direction by one step in the north west south east cycle. A turn of 270 degrees anticlockwise is equivalent to turning 90 degrees clockwise. Once we know his final direction, we can break the path into vertical and horizontal components. The straight line distance between the starting and ending points can then be computed using the Pythagoras theorem as the hypotenuse of a right triangle.

Step-by-Step Solution:
Step 1: The man starts facing east. He first turns towards north. This is a 90 degree anticlockwise turn from east and he walks 5 km in the north direction, reaching (0, 5) if we set the origin at the starting point. Step 2: From facing north, he then turns 270 degrees anticlockwise. A 270 degree anticlockwise rotation is the same as a 90 degree clockwise rotation. From north, a 90 degree clockwise turn makes him face east. Step 3: He now walks 12 km towards the east. Starting from (0, 5), moving 12 km east places him at (12, 5). Step 4: The starting position is (0, 0). The final position is (12, 5). The straight line distance between these two points is the hypotenuse of a right triangle with legs 12 and 5. Step 5: Using Pythagoras theorem, distance^2 = 12^2 + 5^2 = 144 + 25 = 169. Step 6: Therefore, distance = square root of 169 = 13 km.

Verification / Alternative check:
Visually, you can sketch a point, move 5 units up for the northward walk, and then 12 units to the right for the eastward walk. This forms a right triangle with a base of 12 and height of 5. Even without full calculation, you may recall the 5 12 13 Pythagorean triple, which confirms that the hypotenuse is 13 units. This matches the computed value and ensures the calculation is consistent.

Why Other Options Are Wrong:
The options 17 km, 11 km, and 9 km do not satisfy the Pythagoras relationship for the legs of 5 km and 12 km. For example, 11^2 = 121 and 9^2 = 81, both of which do not equal 169 when squared. Only 13 km correctly fits the 5 12 13 relationship and corresponds to the exact distance between (0, 0) and (12, 5).

Common Pitfalls:
Some test takers mistakenly treat a 270 degree anticlockwise turn as if it were 90 degrees anticlockwise, leading to a different direction and wrong coordinates. Others forget to apply the Pythagoras theorem or incorrectly add distances. Recognising that 270 degrees anticlockwise equals 90 degrees clockwise is a key conceptual step here.

Final Answer:
The minimum straight line distance between his starting and ending positions is 13 km.