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.

Answer 1

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