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Question
From the top of a tower 60 m high, the angles of depression of the top and bottom of pole are observed to be 45° and 60° respectively. Find the height of the pole.
Solution
From the adjoining figure, in right-angled ΔBED,
`"DE"/"BE" = tan 45°`
DE = BE .....(1)
In right-angled ΔACD,
`"CD"/"AC" = tan 60°`
`60/"AC" = sqrt3`
AC = `60/sqrt3`
AC = `(60 xx sqrt3)/(sqrt3 xx sqrt3)`
AC = `(60 xx sqrt3)/3`
AC = `20sqrt3`
From (1), DE = BE = AC = `20 sqrt3`
Now, Hight of the tower = CD - DE
= `(60 - 20sqrt3)`m
= `20(3 - sqrt3)`m
= 20(3 − 1.73)
= 20(1.27)
= 25.4 m
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