Question

A PERSON MOVES 20 m TOWARDS NORTH AND THEN TURNS 60 DEGREE EAST OF NORTH AND MOVES 10 m.FIND THE MAGNITUDE AS WELL AS DIRECTION OF DISPLACEMENT.

Answer 1

let us say North direction is positive y axis.  Let us say  East direction is positive x axis.  Let i and j be the unit vectors along x and y directions.

Let the person be at the origin at t = 0.
60 deg. East of North  means,  30 deg. from x-axis.

1st movement:   20  j meters
2nd movement:  10 Cos 30 i + 10 Sin 30  j  meters

net displacement:  add the two:

s = 10 Cos 30 i + (20 + 10 Sin 30) j  meters
   = 5√3  i  + 25  j meters

magnitude of displacement:    √ [ (5√3)² + 25² ]  = 10√7 meters = 26.45 m
direction:    Tan Ф =  25/(5√3)    =>  Ф =  70.893  deg from East towards North.

or,  direction is :  19.10 deg due East from North