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Question
The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building (use `sqrt3`=1.73)
Solution
Let the height of the tower AB be h m and the horizontal distance between the tower and the building BC be x m.
So,
AE=(h−50) m
In ∆AED,
`tan45^@=(AE)/(ED)`
`=>1=(h-50)/x`
⇒x = h−50 .....(1)
In ∆ABC,
`tan^@=(AB)/(BC)`
`=>sqrt3=h/x`
`=>x=sqrt3=h `
Using (1) and (2), we get
`x=sqrt3x-50`
`=>x=(sqrt3-1)=50`
`=>x=(50(sqrt3+1))/2=25xx2.73=68.25m`
Substituting the value of x in (1), we get
68.25 = h −50
⇒ h = 68.25 + 50
⇒h=118.25 m
Hence, the height of tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.
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