The angle of elevation of the top of a tower from a point on the ground which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Let AB be the height of the tower and BC = 30 m be the distance from the foot.
In right △ABC, ∠ACB = 30°
$$\tan 30° = \frac{AB}{BC}$$
$$\frac{1}{\sqrt{3}} = \frac{AB}{30}$$
$$AB = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$$
∴ The height of the tower is $10\sqrt{3}$ m.
Source: Exercise 9.1, Q.4; Chapter 9 — Some Applications of Trigonometry
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Explanation
- Draw a right triangle with the tower as the vertical side (AB), the ground distance as the base (BC = 30 m), and angle of elevation at C = 30°.
- Use tan because you have the opposite side (height) and adjacent side (distance).
- Rationalise $\frac{30}{\sqrt{3}}$ by multiplying numerator and denominator by $\sqrt{3}$ to get $10\sqrt{3}$.
- Examiner expects: diagram description (optional but good), correct trig ratio, clear working, and the final answer in simplified surd form.