(i) To find \(\theta\), use the component of the resultant in the direction OA:

\(8 + 8 \cos \theta = 9\)

Solving for \(\cos \theta\):

\(8 \cos \theta = 1\)

\(\cos \theta = \frac{1}{8}\)

\(\theta = \cos^{-1} \left( \frac{1}{8} \right) \approx 82.8^\circ\)

(ii) To find the magnitude of the resultant, use the cosine rule in the triangle formed by the forces and the resultant:

\(R^2 = 8^2 + 8^2 + 2 \times 8 \times 8 \cos \theta\)

\(R^2 = 64 + 64 + 128 \cos \theta\)

Using \(\cos \theta = \frac{1}{8}\):

\(R^2 = 128 + 128 \times \frac{1}{8}\)

\(R^2 = 128 + 16\)

\(R^2 = 144\)

\(R = \sqrt{144} = 12 \text{ N}\)