How do you find the direction of the resultant vector?

Question

A pedestrian walks East at the rate of 6 km per hour. The wind pushed him NorthWest at a rate of 13 km per hour. Find the direction of the resultant vector. 
(round to the nearest tenth)
I am very confused the problem was copied exactly as it was given to me.

Answer 1

Assuming that by northwest you mean 135° from east, counterclockwise, the components of the resultant vector are

u = (6 + 13cos135°) km/hr = -3.192 km/hr
v = (0 + 13sin135º) km/hr = 9.192 km/hr

√(u² + v²) = √[(-3.192)² + (9.192)²] km/hr ≅ 9.73 km/hr

 

The direction is

 

θ = tan^-1(v/u) = tan^-1(9.192/-3.192) = 109.1°

 

Note:  You have to be careful about the choice of quadrant.  You know it must be in quadrant II.

 

The direction is 109.1° from east, with the angle going counterclockwise.