Question

A man goes 15 meters due west and then 8 meters due north. How far is he from the starting point?

Answer 1

SOLUTION :  

Figure is in the attachment.

Let the initial position of the man be O and his final position be B . The man goes 15 m west and then 8 m due north.Hence , ∆OAB is a right angled ∆ at A. OA = 15 m and AB = 8 m.

By Pythagoras theorem,

OB² = OA² + AB²

OB² = (15)² + (8)²

OB² = 225 + 64 = 289  

OB² = 289

OB =√289  

OB = ± 17 m  

OB = 17 m

[Taking positive square root]

Hence, the man is at a distance of 17 m from the starting point.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

the answer is 17m

it can be easilly solved by pythagorus theorm