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Question
The angles of depression and elevation of the top of a 12m high building from the top and the bottom of a tower are 60° and 30° respectively. Find the height of the tower, and its distance from the building.
Solution
Let AB be the building and CD be the tower.
In ΔABD
`"AB"/"BD" = tan 30^circ`
`12/"BD" = 1/sqrt(3)`
`"BD" = 12sqrt(3)`
In ΔACE
`"AE" = "BD" = 12sqrt(3)`
`"CE"/"AE" = tan 60^circ`
`"CE"/(12sqrt(3)) = sqrt(3)`
`"CE" = 12sqrt(3) xx sqrt(3) = 12 xx 3 = 36`
`"CD" = "CE" + "ED" = 36 + 12 = 48`
So, height of the tower is 48 m and its distance from the building is `12sqrt(3)` m = `12 xx 1.732` m = 20.78 m(approximately)
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