A straight highway leads to the foot of a tower. A man standing on the top of the 75 m high tower observes two cars at angles of depression of $30^\circ$ and $60^\circ$, which are approaching the foot of the tower. If one car is exactly behind the other on the same side of the tower, find the distance between the two cars. (use $\sqrt{3} = 1.73$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Let AB = 75 m be the tower. Let C and D be the two cars on the highway, with D farther from the tower.
Diagram: Tower AB vertical; C and D on the ground with D beyond C from the tower. Angles of depression to C and D are 60° and 30° respectively.
In right △ABC (angle of depression = 60°):
$$\tan 60° = \frac{AB}{BC}$$
$$\sqrt{3} = \frac{75}{BC}$$
$$BC = \frac{75}{\sqrt{3}} = \frac{75\sqrt{3}}{3} = 25\sqrt{3} \text{ m}$$
In right △ABD (angle of depression = 30°):
$$\tan 30° = \frac{AB}{BD}$$
$$\frac{1}{\sqrt{3}} = \frac{75}{BD}$$
$$BD = 75\sqrt{3} \text{ m}$$
Distance between the two cars:
$$CD = BD - BC = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} \text{ m}$$
$$CD = 50 \times 1.73 = \textbf{86.5 m}$$
Source: Chapter 9 (Some Applications of Trigonometry), Exercise 9.1
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Explanation
- Key idea: Angle of depression from the top = angle of elevation from the car to the top (alternate angles with the horizontal). So use tan directly in the right triangle formed by tower height and ground distance.
- The car at 60° is closer (smaller distance BC); the car at 30° is farther (larger distance BD). Distance between cars = BD − BC.
- Always rationalize $\frac{75}{\sqrt{3}}$ to get $25\sqrt{3}$.
- Substitute $\sqrt{3} = 1.73$ only at the final step.
- Draw and label the diagram — it typically carries 1 mark in CBSE.