A child went 90 m in the East to look for his father, then he turned right and…

Direction-sense (distance-and-direction) problems are solved by fixing East and North as the positive x- and y-axes; every 'turn right' or 'turn left' rotates the direction of travel by 90° in a fixed sense. The straight-line distance between the start and end points is then the hypotenuse of a right triangle whose legs are the NET east-west displacement and the NET north-south displacement, found using the Pythagorean theorem: distance = √((net E-W)2 + (net N-S)2).

1. Let the starting point be A. The child walks 90 m East to point B.

2. At B he turns right (facing East, a right turn means he now faces South) and walks 20 m to point C.

3. At C he turns right again (facing South, a right turn means he now faces West) and walks 30 m to point D, his uncle's house.

4. From D he walks 100 m North to point E, where he meets his father.

5. Net east-west displacement: 90 m East − 30 m West = 60 m East.

6. Net north-south displacement: 100 m North − 20 m South = 80 m North.

7. Straight-line distance AE = √(602 + 802) = √(3600 + 6400) = √10000 = 100 m.

Cross-check: 60, 80, 100 is a Pythagorean triple (a scaled-up 3-4-5 right triangle), confirming the arithmetic; adding the four displacement vectors (90 m East, 20 m South, 30 m West, 100 m North) directly also resolves to the same net vector of 60 m East and 80 m North.

So the child met his father 100 m from the starting point.