a man walks 8km North and then 5km in a. direction 60degrees. find its distance from his starting point

Answer 1

Use the component method to break down each vector into x,y components.
8km North = <0,8>
5km at 60° = <5cos60°,5sin60°>
Assuming angle is in Cartesian plane (counterclockwise from east) and not bearing (clockwise from north).

Add the two vectors and calculate the magnitude of the resultant.

Answer 2

56

Answer 3

5km

Answer 4

d = 8km[90o] + 5km[60o],
d = X + Yi = (8*Cos90+5*Cos60) + (8*sin90+5*sin60)I = 2.5 + 12.3i,
d = sqrt(X^2+Y^2) = 12.6 km.
Tan A = Y/x.