A man is facing North direction. He turns 90 degrees in the anti-clockwise…

Concept: A facing direction can be written as a bearing measured clockwise from North, so that North = 0°, East = 90°, South = 180° and West = 270°. A clockwise turn adds to the bearing and an anti-clockwise turn subtracts from it. The facing after any sequence of turns is therefore the starting bearing plus the algebraic sum of all the turns, reduced modulo 360°.

Applying the rule to the three turns given in the question:

1. The man starts facing North, so his starting bearing is 0°.

2. The first turn is 90° anti-clockwise, which subtracts: 0° − 90° = − 90°, and − 90° + 360° = 270°. A bearing of 270° is West, so he is now facing West.

3. The second turn is 135° in the same anti-clockwise sense, which subtracts again: 270° − 135° = 135°. A bearing of 135° is South-East, so he is now facing South-East.

4. The third turn is 270° clockwise, which adds: 135° + 270° = 405°, and 405° − 360° = 45°. A bearing of 45° is North-East.

Cross-check: the turns can also be added first, before converting to a direction. Taking anti-clockwise as negative and clockwise as positive, − 90° − 135° + 270° = + 45°, a net turn of 45° clockwise from the starting facing of North, which is again North-East. A second check uses equivalence of turns: a 270° clockwise turn is the same as a 90° anti-clockwise turn, so from South-East (135°) the last turn gives 135° − 90° = 45°. Both routes agree, so the man is finally facing North-East.