trigonometry
posted by karl .
A barge is being towed south at the rate 2 km/hr. A man on the deck walks from west to east at the rate 4ft/s. Find the magnitude and direction of the man's actual velocity.

1 km = appr 3280.84 feet
2 km/h = 2(3280.84/3600) ft/s
= 1.82269 ft/s
let r be the resultant velocity
r^2 = 2^2 + 1.82269^2
r = √7.32219..
= appr 2.7 ft/s
tanØ = 1.82269/2 , where Ø is in my rightangled triangle
Ø = appr 42.34°
his direction is S 42.34° E
or by vectors
vector r = 2(cos270 , sin270) + 1.82269(cos0 , sin0)
= 2(0, 1) + 1.82269(1,0)
= (0,2) + (1.82269,0)
= (1.82269,2)
magnitude = √(1.82269^2 + (2)^2)
= 2.7 as above
direction:
tan k = 2/1.82269
k = 312.34° which is the same as S 42.34 W 
X = 4 Ft/s
Y = 2km/h = 2000m/3600s * 3.3Ft/m = 1.833 Ft/s
Tan A = Y/X = 1.833/4 = 0.45825
A = 24.62o = 24.62o South of East. =
65.38o East of South = Direction.
Magnitude = X/Cos A = 4/Cos(24.62) =
4.4 Ft/s.