Resolve the forces into horizontal ( X ) and vertical ( Y ) components:

Horizontal component ( X ):

\(X = 20 \, \cos 60° - 14 - 16 \, \cos 50° = -14.2846 \, \text{N}\)

Vertical component ( Y ):

\(Y = 60 - 20 \, \sin 60° - 16 \, \sin 50° = 30.42278 \, \text{N}\)

Calculate the magnitude of the resultant force ( R ):

\(R = \sqrt{(-14.2846)^2 + (30.42278)^2} = 33.6 \, \text{N}\)

Calculate the direction of the resultant force:

\(\theta = \tan^{-1}\left(\frac{30.42278}{14.2846}\right) = 64.8°\)

The direction is 64.8° above the 14 N force.