Question
The angle of elevation of the top of a building from a
point on the ground is 30° and moving 40 meters towards the building it becomes 60°. The height of the building is?Solution
Let AB bee the height of the building. In ∆ACD and ∆ABC ∠ACB = ∠CAD + ∠CDA 60° = ∠CAD + 30° ∠CAD = 30° So, AC = CD AC = 40 m cosec 60° = AC/AB 2/√3 = 40/AB AB = 20√3
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