A drone was used to facilitate movement of an ambulance on the straight highway to a point P on the ground where there was an accident. The ambulance was travelling at the speed of 60 km/h. The drone stopped at a point Q, 100 m vertically above the point P. The angle of depression of the ambulance was found to be 30° at a particular instant.
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding stimulus
Model Answer
(i) Diagram:
```
Q
| \
| \
100m | \ (angle of depression = 30°)
| \
P__________A
```
Q is directly above P (QP = 100 m), A is the ambulance on the highway.
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(ii) Let A be the position of the ambulance. In △QPA:
$$\tan 30° = \frac{QP}{PA} = \frac{100}{PA}$$
$$PA = \frac{100}{\tan 30°} = 100\sqrt{3} = 100 \times 1.73 = \textbf{173 m}$$
---
(iii) When angle of depression = 45°, let the new position be A′:
$$\tan 45° = \frac{100}{PA'} \Rightarrow PA' = 100 \text{ m}$$
Distance covered by ambulance = PA − PA′ = 173 − 100 = 73 m
Speed = 60 km/h $= 60 \times \dfrac{1000}{3600} = \dfrac{50}{3}$ m/s
$$\text{Time} = \frac{73}{\frac{50}{3}} = \frac{73 \times 3}{50} = \frac{219}{50} = \textbf{4.38 seconds}$$
Explanation
- (i) A right-angle triangle with vertical side QP = 100 m and horizontal side PA is expected. Even a rough labelled sketch scores the mark.
- (ii) Use tan(angle) = opposite/adjacent. The angle of depression from Q equals the angle of elevation from A (alternate interior angles), so tan 30° = 100/PA directly.
- (iii) Find PA′ using tan 45° = 1, subtract to get distance covered, convert speed to m/s, then apply Time = Distance/Speed. Show every step clearly for full 2 marks.