(a) (M1A1A1A1, N4).

Working:
(i)
Using the sine rule: \(\frac{25}{\sin 28^\circ} = \frac{50}{\sin \angle OBA}\) (M1).
\(\sin \angle OBA = \frac{50 \sin 28^\circ}{25} = 2 \sin 28^\circ\) (A1).
\(\sin 28^\circ \approx 0.4695\), so \(\sin \angle OBA \approx 0.939\) (A1).
\(\angle OBA \approx \arcsin(0.939) \approx 69.87^\circ\) or \(180^\circ – 69.87^\circ \approx 110.13^\circ\) (A1).
\(\angle OAB = 180^\circ – 28^\circ – \angle OBA\), so \(\angle OAB \approx 82.13^\circ\) or \(41.87^\circ\) (rounded to \(82.1^\circ\) or \(41.9^\circ\)).

(a) (M1A1A1, N3).

Working:
(ii)
Area = \(\frac{1}{2} \times 50 \times 25 \times \sin \angle OAB\) (M1).
For \(\angle OAB \approx 82.1^\circ\), \(\sin 82.1^\circ \approx 0.9903\), area \(\approx 625 \times 0.9903 \approx 619 \, \text{m}^2\) (A1).
For \(\angle OAB \approx 41.9^\circ\), \(\sin 41.9^\circ \approx 0.6691\), area \(\approx 625 \times 0.6691 \approx 417 \, \text{m}^2\) (A1).
The two possible areas are \(619 \, \text{m}^2\) and \(417 \, \text{m}^2\).

(b) (M1A1, N2).

Working:
Area \(60 = \frac{1}{2} \times x \times y \times \sin 28^\circ\) (M1).
\(120 = x y \sin 28^\circ\), so \(x y = \frac{120}{\sin 28^\circ}\) (A1).
Cosine rule: \(10^2 = x^2 + y^2 – 2 x y \cos 28^\circ\), so \(x^2 + y^2 = 100 + 2 x y \cos 28^\circ\).
Since \(2 x y \cos 28^\circ = \frac{2 \times 120 \cos 28^\circ}{\sin 28^\circ} = \frac{240}{\tan 28^\circ}\), thus \(x^2 + y^2 = 100 + \frac{240}{\tan 28^\circ}\).

(c) (M1A1, N2).

Working:
From (b), \(x^2 + y^2 = 100 + \frac{240}{\tan 28^\circ} \approx 551.38\) (M1).
Area \(60 = \frac{1}{2} \times x \times y \times \sin 28^\circ\), so \(x y \approx 255.59\).
Solve \(x^2 + \frac{(255.59)^2}{x^2} = 551.38\), leading to \(x^4 – 551.38 x^2 + 65327 \approx 0\).
Quadratic in \(u = x^2\): \(u \approx 379\) or \(172\), so \(x \approx 19.5 \, \text{m}\) or \(13.1 \, \text{m}\) (A1).
The two possible lengths of OC are \(13.1 \, \text{m}\) and \(19.5 \, \text{m}\).

Total [11 marks]