Question:
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is \(30^\circ\). After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is \(60^\circ\). Then the time taken (in minutes) by him, from B to reach the pillar, is :
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is \(30^\circ\). After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is \(60^\circ\). Then the time taken (in minutes) by him, from B to reach the pillar, is :
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Express the distances from A and from B to the base of the pillar in terms of the height \(h\).
Updated On: Nov 11, 2026
- 6
- 10
- 20
- 5
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The Correct Option is D
Solution and Explanation
Let the pillar height be \(h\) and its base be \(O\). From A: \(\tan30^\circ=\frac{h}{OA}\Rightarrow OA=h\sqrt3\). From B: \(\tan60^\circ=\frac h{OB}\Rightarrow OB=\frac{h}{\sqrt3}\).
\(AB=OA-OB=h\sqrt3-\frac h{\sqrt3}=\frac{2h}{\sqrt3}\), covered in 10 minutes.
Speed \(=\frac{2h/\sqrt3}{10}\) per minute. Time from B to O \(=\frac{h/\sqrt3}{(2h/\sqrt3)/10}=\frac{10}{2}=5\) minutes.
Answer: (4) 5
\(AB=OA-OB=h\sqrt3-\frac h{\sqrt3}=\frac{2h}{\sqrt3}\), covered in 10 minutes.
Speed \(=\frac{2h/\sqrt3}{10}\) per minute. Time from B to O \(=\frac{h/\sqrt3}{(2h/\sqrt3)/10}=\frac{10}{2}=5\) minutes.
Answer: (4) 5
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