From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Diagram: Let AB = 20 m (building), BC = tower, P = point on ground, PB = horizontal distance.
Step 1: In △PAB, angle of elevation of bottom of tower (top of building) = 45°.
$$\tan 45° = \frac{AB}{PA} \implies 1 = \frac{20}{PA} \implies PA = 20 \text{ m}$$
Step 2: Let height of tower BC = h m. Then total height PC = (20 + h) m.
In △PAC, angle of elevation of top of tower = 60°.
$$\tan 60° = \frac{AC}{PA} \implies \sqrt{3} = \frac{20 + h}{20}$$
Step 3: Solving:
$$20\sqrt{3} = 20 + h$$
$$h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \text{ m}$$
∴ Height of the tower = $20(\sqrt{3} - 1)$ m ≈ 14.64 m
Source: Exercise 9.1, Q.7; Chapter 9 — Some Applications of Trigonometry
---
Explanation
- The key is setting up two separate right triangles from the same point P: one to the base of the tower (top of building) and one to the top of the tower.
- Since 45° elevation gives tan 45° = 1, the horizontal distance equals the building height (20 m) — this is the crucial first step.
- Examiners expect a labelled diagram, both tan equations shown clearly, and the final answer simplified to $20(\sqrt{3}-1)$ m. Writing the decimal approximation is optional but appreciated.