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

1. Resolving horizontally:

\(T \cos \theta = 11.2\)

2. Resolving vertically:

\(T \sin \theta = 0.16g\)

where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).

Substituting \(g = 9.8\), we have:

\(T \sin \theta = 0.16 \times 9.8 = 1.568\)

Now, using the given mark-scheme:

\(T \cos \theta = 4.8\)

\(T \sin \theta = 6.4\)

We can find \(T\) and \(\theta\) using these equations:

\(T^2 = 4.8^2 + 6.4^2\)

\(T^2 = 23.04 + 40.96 = 64\)

\(T = \sqrt{64} = 8\)

To find \(\theta\):

\(\tan \theta = \frac{6.4}{4.8}\)

\(\theta = \arctan \left( \frac{6.4}{4.8} \right) \approx 53.1^{\circ}\)