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Question
From the top of the tower 60 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post (tan 38° = 0.7813, `sqrt(3)` = 1.732)
Solution
Let the height of the lamp post be h
The height of the tower (BC) = 60 m
∴ EC = 60 – h
Let AB be x
In the right ∆ABC,
tan 60° = `"BC"/"AB"`
`sqrt(3) = 60/x`
x = `60/sqrt(3)` ...(1)
In the right ∆DEC, tan 38° = `"EC"/"DE"`
0.7813 = `(60 - "h")/x`
x = `(60 - "h")/(0.7813)` ...(2)
From (1) and (2) we get
`60/sqrt(3) = (60 - "h")/(0.7813)`
60 × 0.7813 = `60 sqrt(3) = sqrt(3)`h
`sqrt(3)"h" = 60 sqrt(3) - 46.88`
= 60 × 1.732 – 46.88
= 103.92 – 46.88
1.732h = 57.04
⇒ h = `(57.04)/(1.732)`
h = `(570440)/(1732)`
= 32.93 m
∴ Height of the lamp post = 32.93 m
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