Question

From the top of a building, the angle of elevation of the top of a cell tower is 60º and the angle of depression to its foot is 45º. If distance of the building from the tower is 7m, then find the height of the tower.

Answer 1

Height of the tower BE = BC + CE = h m

Height of the building AD = BC = x m

Distance between foot of the tower and

foot of building AB = 7 m

CE = BE - BC = ( h - x ) m

Angle of elevation from D = <CDE = 60°

Angle of depression from D = <CDB = 45°

Therefore ,

<ABD = <CDB = 45° ( alternate angles )

In ∆CDE ,

tan 60° = CE/CD = ( h - x )/7

=> √3 = ( h - x )/7

( h - x ) = 7√3 ----( 1 )

In ∆ABD ,

tan 45° = AD/AB

1 = x/7

x = 7 -----( 2 )

From ( 1 ) and ( 2 ) ,

h - 7 = 7√3

h = 7√3 + 7

h = 7( √3 + 1 )

h = 7( 1.732 + 1 )

h = 7 × 2.732

h = 19.124 m

Therefore ,

Height of the tower = h = 19.124 m

I hope this helps you.

: )