A man walks `10 m` towards east and then turns at an angle of `30^(0)` to the north of east and walks 10 m`. Calculate the net displacement of the man. Also find the direction of net displacement.

Explanation

Refer to fig. `2 . (c ) .45. the net displacement of man
.
`=vec (OB) =vec (OA) + vec (AB)`
wher ` OA=10 m =AB` and angle betwee ` vec (OA)` and ` vec (OB)` is 30^(@)`
Magnitude of ` vec (OB)` is
` | vec (OB)| =sqrt ((OA)^(2) +(AB)^(2) + 2 (OA) (AB) cos 30^(@))`
`= sqrt (10^(2) + 10^(2) + 2 xx 10 xx 10 xx sqrt 3//2)`
`= 10 sqrt (1 +1 +1 1.732) =10 sqrt 3.732)`
`=10 xx 1.93 =19 .3 m`
Direction fo `vec (OB)`
` tan beta =(10 sin 30 ^(@))/(10 + 10 cos 30^(@)) =(10 xx 1//2 )/(10 + 10 xx sqrt 3//2)`
`1/(2+ sqrt3) = 1/(2 + 1.732) 1/(3.732) =0.26 79`
`=tan `15^(@)`
or `beta =15 ^(2)`.