Question
From the top of a cliff 100 meter high, the angles of
depression of the top and bottom of a tower are 45° and 60° respectively. The height of the tower isSolution
Let height of tower, be x in ∆AEC => tan 45° = AE / CE => 1 = (100 – x) / CE => CE = 100-x ∆ABD tan 60° = AB / BD => √3 = 100 / CE (CE = BD) => (100 – x) × √3 = 100 => 100 – x = (100 / √3) => x = 100 – 100 / √3 => x = 100 (1 – 1 / 100 / √3) => x = 100/ 3 (3 – √3)
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