\(F = F\cos \theta + F\sin \theta\) (Resolving into its components)

\(10N = 10 \cos 60, 10 \sin 60\)

\(6N = 6 \cos 330, 6 \sin 330\)

\(R = F_{1} + F_{2} = (10 \cos 60, 10 \sin 60) + (6 \cos 330, 6 \sin 330)\)

The y- component : \(10 \sin 60 + 6 \sin 330 = 10 \times \frac{\sqrt{3}}{2} + 6 \times \frac{-1}{2}\)

= \((5\sqrt{3} - 3)N\)