Two pillars of equal lengths stand on either side of a road which is 100 m wide, exactly opposite to each other. At a point on the road between the pillars, the angles of elevation of the tops of the pillars are 60° and 30°. Find the length of each pillar and distance of the point on the road from the pillars. (Use $\sqrt{3} = 1.732$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Diagram: Let AB and CD be two pillars of equal height h m. Let P be the point on the road between them, with BP = x m, so PD = (100 − x) m.
In △ABP (angle of elevation = 60°):
$$\tan 60° = \frac{AB}{BP} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \cdots (1)$$
In △CDP (angle of elevation = 30°):
$$\tan 30° = \frac{CD}{PD} \implies \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \implies h = \frac{100 - x}{\sqrt{3}} \quad \cdots (2)$$
From (1) and (2):
$$x\sqrt{3} = \frac{100 - x}{\sqrt{3}}$$
$$3x = 100 - x \implies 4x = 100 \implies x = 25 \text{ m}$$
Height of each pillar:
$$h = 25\sqrt{3} = 25 \times 1.732 = 43.3 \text{ m}$$
Distances:
- Distance from pillar AB = BP = 25 m
- Distance from pillar CD = PD = 100 − 25 = 75 m
∴ The length of each pillar is 43.3 m, and the point P is 25 m from one pillar and 75 m from the other.
Source: Chapter 9, Exercise 9.1 (Q.10 variant), Heights and Distances
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Explanation
- Set up two right triangles sharing the point P; label distances as x and (100 − x).
- Use tan for both elevation angles since you have opposite (height) and adjacent (horizontal distance).
- Equate the two expressions for h to find x, then substitute back.
- Examiners award marks for: diagram/setup (1), correct equations (1+1), solving for x (1), final height with √3 value (1).
- Note: the pillar with the larger elevation angle (60°) is closer to P — a quick sanity check.