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A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards it, the tower is found to subtend an angle twice as before. The height of the tower is
80m
100m
160m
200m
Given, BC = 160 m and CD = 100 m ∴ BD = BC – CD = 160 – 100 = 60 m Let, the height of the tower, AB = h metre And, ∠ACB = Θ and ∠ADB = 2Θ In ΔABC,
By: Amit Kumar ProfileResourcesReport error
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