Question

A man observes the angle of elevation of the top of the tower to be 45°. He walks towards it in a horizontal line through its base. On covering 20 m the angle of elevation changes to 60°. Find the height of the tower correct to 2 significant figures. 

Answer 1

Let the height of the tower be ‘h’ m. 

In Δ ADC , tan 45° = `h/(20 + x)`

                         1 = `h /(20+ x`

⇒ h = 20 + x 

Also , In ΔBDC , tan 60° = `h/x`

                               `sqrt(3) = h / x`

                               ⇒ x = `h/(sqrt(3))`              ...(2)

h = 20 + `h/sqrt(3)`

`h - h/sqrt(3) = 20`

`h((sqrt(3) -1)/sqrt(3)) = 20`

`h = (20 sqrt(3)) /((sqrt(3) - 1)) xx ((sqrt(3) + 1)) /(( sqrt(3) +1 )`

`= (20 (3 + sqrt(3)))/(3-1)`

`= (20 (3+1.732))/2`

= 10 (4.732)

Height of the tower . h = 47.32 m