From the top of a 7 m high building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Determine the height of the tower.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Let AB = 7 m (building), CD = height of cable tower, BD = horizontal distance between them.
Draw AE ∥ BD, so AE = BD and ED = AB = 7 m.
Step 1: Find BD using angle of depression (30°)
From A, angle of depression of foot D = 30°.
In △ABD, ∠ADB = 30° (alternate angles):
$$\tan 30° = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{7}{BD} \implies BD = 7\sqrt{3} \text{ m}$$
So AE = $7\sqrt{3}$ m.
Step 2: Find CE using angle of elevation (60°)
From A, angle of elevation of top C = 60°.
In △AEC:
$$\tan 60° = \frac{CE}{AE} \implies \sqrt{3} = \frac{CE}{7\sqrt{3}} \implies CE = 7\sqrt{3} \times \sqrt{3} = 21 \text{ m}$$
Step 3: Total height of tower
$$CD = CE + ED = 21 + 7 = \boxed{28 \text{ m}}$$
The height of the cable tower is 28 m.
Source: Chapter 9, Section 9.1 (Heights and Distances)
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Explanation
- The key construction is drawing AE horizontal (parallel to BD), splitting the tower height into two parts: ED = 7 m (equal to building height) and CE (found using tan 60°).
- Angle of depression = 30° gives the horizontal distance BD via tan 30°; this same distance is AE used in the second triangle.
- Examiners award marks for: correct diagram (1), correct use of tan 30° to find BD (1–2), correct use of tan 60° to find CE (1–2), and final addition (1). Always show both triangles separately and write the final answer clearly.
- Note: The textbook (Q.12 in Exercise 9.1) has 45° for angle of depression, giving a different answer (7+7√3 m). This question uses 30°, so the answer is 28 m.