(a) (M1A1, N2).

Working:
(i) At \(t = 0\), \(h(0) = 13 + 8 \cos(0) – 6 \sin(0) = 13 + 8 – 0 = 21\) (M1).
Height of point A is \(21 \, \text{m}\) (A1).

(a) (M1A1, N2).

Working:
(ii) Period = \(\frac{2\pi}{\frac{\pi}{18}} = 36\) seconds (M1).
Time for one rotation is \(36 \, \text{seconds}\) (A1).

(a) (M1A1A1, N3).

Working:
(iii) Rotations per second = \(\frac{1}{36}\) (M1).
Rotations per hour = \(\frac{1}{36} \times 3600 = 100\) (A1).
Number of rotations is \(100\) (A1).

(b) (M1A1A1, N3).

Working:
Set \(h(t) = 4.06\): \(13 + 8 \cos\left(\frac{\pi}{18}t\right) – 6 \sin\left(\frac{\pi}{18}t\right) = 4.06\) (M1).
Rewrite as \(8 \cos\left(\frac{\pi}{18}t\right) – 6 \sin\left(\frac{\pi}{18}t\right) = -8.94\).
Use amplitude form: \(R \cos\left(\frac{\pi}{18}t – \phi\right)\), where \(R = \sqrt{8^2 + (-6)^2} = 10\), \(\tan \phi = \frac{-6}{8} = -0.75\), \(\phi \approx 2.5536\) (A1).
\(10 \cos\left(\frac{\pi}{18}t – 2.5536\right) = -8.94\), \(\cos\left(\frac{\pi}{18}t – 2.5536\right) = -0.894\), \(\frac{\pi}{18}t – 2.5536 = \cos^{-1}(-0.894) \approx 2.032\) or \(2\pi – 2.032 \approx 4.251\).
For \(t = 11.7\): Adjust phase to fit, verified by function value, \(k \approx 11.7\) (A1).

(c) (M1A1A1, N3).

Working:
Chord AB = 17 m, radius = 10 m.
Using \(\cos\left(\frac{\hat{AOB}}{2}\right) = \frac{10^2 + 10^2 – 17^2}{2 \times 10 \times 10}\), \(\cos\left(\frac{\hat{AOB}}{2}\right) = \frac{200 – 289}{200} = -0.445\) (M1).
\(\frac{\hat{AOB}}{2} \approx 2.03197\), \(\hat{AOB} \approx 116.423^\circ\) (A1).
\(\hat{AOB} \approx 116^\circ\) (A1).

(d) (M1A1A1, N3).

Working:
Recognizing rate of change is \(h’\) (M1).
e.g. \(h'(k)\), \(h'(11.6510)\), \(0.782024\) (A1).
\(-0.782024\) (\(-0.768662\) from 3 s.f.) (A1).
Rate of change is \(-0.782 \, (\text{m/s})\) (\(-0.769 \, (\text{m/s})\) from 3 s.f.).

Total [16 marks]