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Question
The angles of depression of the top and bottom of a tower as seen from the top of a 60 `sqrt(3)` m high cliff are 45° and 60° respectively. Find the height of the tower.
Solution
Let AD be the tower and BC be the cliff.
We have,
BC = 60 `sqrt(3)` , ∠ CDE = 45° and ∠BAC = 60°
Let AD = h
⇒ BE = AD = h
⇒ CE = BC - BE= 60 `sqrt(3)` - h
In ΔCDE,
` tan 45° = (CE)/(DE)`
`⇒ 1 = (60 sqrt(3) -h)/(DE)`
`⇒ DE = 60 sqrt(3) - h`
`⇒ AB = DE = 60 sqrt(3) - h` ............(1)
Now, in ΔABC
`tan 60° = (BC)/(AB)`
`⇒ sqrt(3)= (60 sqrt(3) )/ (60 sqrt(3) -h)` [ Using (1)]
`⇒ 180 - h sqrt(3) = 60 sqrt(3)`
`⇒ h sqrt(3) = 180- 60 sqrt(3)`
`⇒ h = (108 -60sqrt(3) )/sqrt(3) xx sqrt(3)/sqrt(3)`
`⇒ h = ( 180 sqrt(3)-180) /3`
`⇒ h = (180 sqrt(3)-1) /3`
∴` h = 60 ( sqrt(3)-1)`
= 60 (1.732 -1)
= 60 (0.7.32)
Also, h = 43.92m
So, the height of the tower is 43. 92 m.
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