EXAMPLE 4 : Determine the resultant of 26N,
10^{0}, 40N, 115^{0}, 63N, 188^{0}, 60N, 260^{0}
Solution
By the Law of Polygon, place tail of each vector in turns at the arrow of the preceding vector.
= Resultant
\  = 70 N with q_{x} = 203^{0 } (obtained from the measurement of the figure)
EXAMPLE 5 : A car is traveling due east with the speed of 180 miles per hour. There is a storm, blowing in North west direction at 60 miles per hour. Draw a diagram that represents the car's speed and direction relative to the storm.
Solution
Let V be the velocity vector for car speed with  v  = 180 miles per hour due east.
Let u be the velocity vector for storm speed with u = 60 miles per hour due North west
The bearing of v is the angle measured from due north to v is
270^{0 }and
The bearing of u is the angle measured clockwise from due north
150
Using tail up rule, the length and bearing of the resultant calculated as by using Laws of
Cosine as
R^{2} = u^{2} + v^{2}  2uv cos 45^{0 }
R^{2} = ( 60 )^{2} + (180)^{2}  2 (60) (180) cos 45^{0}
R^{2} = 3600 + 32400  2 (21600) (0.7071)
R^{2} = 20726.4935 \ R = 143.97 miles per hour.
Now using Sine Law,
Thus the bearing angle = (90 17.14)^{0} = 72.86^{0}
