In Fig. 3, AB is tower of height 50 m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the two cars.
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding rag
Model Answer
Let AB = 50 m be the tower. Let C and D be the two cars on opposite sides.
For car C (angle of depression = 30°):
In △ABC, ∠ACB = 30° (alternate angles)
$$\tan 30° = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{50}{BC}$$
$$BC = 50\sqrt{3} \text{ m}$$
For car D (angle of depression = 45°):
In △ABD, ∠ADB = 45°
$$\tan 45° = \frac{AB}{BD} \implies 1 = \frac{50}{BD}$$
$$BD = 50 \text{ m}$$
Distance between the two cars:
$$CD = BC + BD = 50\sqrt{3} + 50 = 50(\sqrt{3} + 1) \text{ m}$$
Source: Chapter 9, Heights and Distances
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Explanation
- The angle of depression equals the alternate interior angle at the base (since the horizontal from the top is parallel to the ground). Always state this step — examiners award a mark for it.
- Use tan (opposite/adjacent) since height and base distance are involved.
- Final answer must be left as $50(\sqrt{3}+1)$ m or approximately 136.6 m. Both forms are acceptable; exact form is preferred.
- 3 marks are typically split: 1 mark for correct triangle/setup, 1 mark for BC, 1 mark for BD and final answer.