From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of 30 m high building are 30° and 60°, respectively. Find the height of the transmission tower. (Use $\sqrt{3} = 1{\cdot}73$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Diagram: Let P be the point on the ground, AB = 30 m be the building, and BC = h be the transmission tower. Let PA = d.
Step 1: In △PAB (angle of elevation to bottom of tower = 30°):
$$\tan 30° = \frac{AB}{PA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{d}$$
$$d = 30\sqrt{3} \text{ m}$$
Step 2: In △PAC (angle of elevation to top of tower = 60°, AC = AB + BC = 30 + h):
$$\tan 60° = \frac{AC}{PA} \Rightarrow \sqrt{3} = \frac{30 + h}{30\sqrt{3}}$$
$$30 + h = \sqrt{3} \times 30\sqrt{3} = 90$$
$$h = 90 - 30 = 60 \text{ m}$$
Therefore, the height of the transmission tower = 60 m.
(Using $\sqrt{3} = 1.73$, the distances are consistent; the tower height is exactly 60 m.)
Source: Chapter 9, Exercise 9.1 (Q.7 variant), Heights and Distances
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Explanation
- The key is setting up two separate right triangles from the same point P: one to the base of the tower (bottom of tower = top of building) and one to the top of the tower.
- Use tan for both since you have opposite (height) and adjacent (horizontal distance) sides.
- Find the horizontal distance from the first triangle, then substitute into the second.
- Examiners award marks for: correct diagram (1 mark), correct equation for tan 30° (1 mark), correct equation for tan 60° (1 mark), solving for d (1 mark), and final answer h = 60 m (1 mark).