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Question
From the top of a cliff, 60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.
Solution
Let AB be the cliff and CD be the tower.
Here AB = 60 m, ∠ADE = 30° and ∠ACB = 60°
In ΔABC,
`(AB)/(BC) = tan 60^circ = sqrt(3)`
`=> BC = (60)/(sqrt(3))`
In ΔADE,
`(AE)/(DE) = tan 30^circ`
`=>` AE = DE tan 30°
= `60/(sqrt(3)) xx 1/sqrt(3)` ...[∵ DE = BC]
= 20 m
∴ CD = EB
= AB – AE
= (60 – 20)
= 40 m
Hence, height of the tower is 40 m.
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