To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

(i) Resolving vertically:

The vertical components of the tension must balance the weight of the ring. Let T be the tension in the string.

From the diagram, the vertical components are:

\(T \sin 50^\circ\) and \(T \sin 20^\circ\)

These must sum to the weight of the ring:

\(T \sin 50^\circ = T \sin 20^\circ + 0.8g\)

Solving for T using \(g = 9.8 \text{ m/s}^2\):

\(T \sin 50^\circ - T \sin 20^\circ = 0.8 \times 9.8\)

\(T(\sin 50^\circ - \sin 20^\circ) = 7.84\)

\(T = \frac{7.84}{\sin 50^\circ - \sin 20^\circ}\)

Calculating gives \(T \approx 18.9 \text{ N}\).

(ii) Resolving horizontally:

The horizontal components of the tension must balance the horizontal force X.

\(X = T \cos 50^\circ + T \cos 20^\circ\)

Substitute \(T = 18.9 \text{ N}\):

\(X = 18.9 \cos 50^\circ + 18.9 \cos 20^\circ\)

Calculating gives \(X \approx 29.9 \text{ N}\).