(a)
In triangle HAC, right-angled at A, \( HA = 380 \, \text{m} \), \( \angle HCA = 25^\circ \)
Use tangent: \( \tan 25^\circ = \frac{380}{AC} \), so \( AC = \frac{380}{\tan 25^\circ} \) (M1)
\( \tan 25^\circ \approx 0.4663 \), \( AC \approx \frac{380}{0.4663} \approx 814.912 \, \text{m} \approx 815 \, \text{m} \) (A1)
Result: 815 m (814.912…) [2]

(b)
METHOD 1
In triangle HAB, right-angled at A, \( HA = 380 \, \text{m} \), \( \angle HBA = 40^\circ \)
Use tangent: \( AB = \frac{380}{\tan 40^\circ} \), \( \tan 40^\circ \approx 0.8391 \), so \( AB \approx \frac{380}{0.8391} \approx 452.866 \, \text{m} \) (M1)
Points C, B, A collinear, so \( AC = AB + BC \), thus \( BC = AC – AB \approx 814.912 – 452.866 \approx 362.046 \, \text{m} \approx 362 \, \text{m} \) (A1)
METHOD 2
In triangle HBC, \( \angle HCB = 25^\circ \), \( \angle HBC = 40^\circ \), so \( \angle BHC = 180^\circ – 25^\circ – 40^\circ = 115^\circ \)
Use sine rule: \( HB = \frac{380}{\sin 40^\circ} \approx \frac{380}{0.6428} \approx 591.175 \, \text{m} \) (M1)
\( \frac{BC}{\sin 15^\circ} = \frac{591.175}{\sin 25^\circ} \), so \( BC \approx \frac{591.175 \times \sin 15^\circ}{\sin 25^\circ} \approx 362.046 \, \text{m} \approx 362 \, \text{m} \) (A1)
Result: 362 m (362.046…) [2]

(c)
Distance \( BC \approx 362.046 \, \text{m} \), time = 15 minutes = \( \frac{15}{60} = 0.25 \, \text{hours} \) (M1)
Speed = \( \frac{\text{distance}}{\text{time}} = \frac{362.046}{0.25} \approx 1448.18 \, \text{m/h} \approx 1450 \, \text{m/h} \) (A1)
Result: 1450 m/h (1448.18…) [2]