Question

A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

Answer 1

Let AB be the tower of height h. And person makes an angle of elevation of the top of a tower is 30°, he walks 50 m towards the foot of tower then makes an angle of elevation of 60°

Let BC = x, CD = 50, and ∠ACB = 60°, ∠ADB = 30°

Now we have to find the height of the tower.

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle ABC

`=> tan C = (AB)/(BC)`

`=> tan 60^@ = h/x`

`=> sqrt3 = h/x`

`=> x = h/sqrt3`

Again in a triangle ADB,

`=> tan D = (AB)/(BC + CD)`

`=> tan 30^@ = h/(x + 50)`

`=> 1/sqrt3 = h/(x + 50)`

`=> sqrt3h = x + 50`

`=> 3h = h + 50sqrt3`

`=> 2h = 50sqrt3`

`=> h = 25sqrt3`

=> h = 25 x 1.73

=> h = 43.25

Hence the height of tower is 43.25 m