The angle of elevation of a tower from a certain point of bus stand is 30°. When a man walks 5m ahead in the direction of the tower, the angle of elevation becomes 60°. What is the height of the tower?
Let, the height of the post = AB = ‘h’ m From ΔACB, ∠ACB = 30° [∵ given] tan ∠ACB = perpendicular/base ⇒ tan 30° = AB/BC ⇒ 1/√3 = h/BC ⇒ BC = h√3 ... (1) From ΔADB, ∠ADB = 60° [∵ given] tan ∠ADB = perpendicular/base ⇒ tan 60° = AB/BD ⇒ √3 = h/BD ⇒ BD = h/√3 ... (2) According to the question: BC - BD = 5 ⇒ h√3 - h/√3 = 5 ⇒ 2h/√3 = 5 ⇒ h = (5√3)/2 m
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