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Question
The angles of elevation and depression of the top and bottom of a lamp post from the top of a 66 m high apartment are 60° and 30° respectively. Find the distance between the lamp post and the apartment `(sqrt(3) = 1.732)`
Solution
Let the height of the lamp post AE be h m
DE = h – 66
Let AB be x
In the right ∆ABC, tan 30° = `"BC"/"AB"`
`1/sqrt(3) = 66/x`
x = `66sqrt(3)` ...(1)
In the right ∆CDE, tan 60° = `"DE"/"DC"`
`sqrt(3) = ("h" - 66)/x`
⇒ `sqrt(3)x` = h – 66
x = `("h" - 66)/sqrt(3)` ...(2)
From (1) and (2) we get
`("h" - 66)/sqrt(3) = 66sqrt(3)`
h – 66 = `66sqrt(3) xx sqrt(3)` = 66 × 3
h – 66 = 198 ⇒ h = 198 + 66
h = 264 m
Distance between the lamp post and the apartment
= `66 sqrt(3)` m
= 66 × 1.732
= 114.31 m
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