Question

The length of the shadow of a tower standing on the level plane is found to 2x meter longer when the sun's altitude is 30° than when it was 45°. Prove that the height of the tower is `x(sqrt3 + 1)` meters.

Answer 1

Let AB be the tower of height hm. the length of the shadow of the tower to be found 2xmeters at the plane longer when sun’s altitude is 30° than when it was 45°.

Let BC = ym,

CD = 2x m and ∠ADB = 30°, ∠ACB = 45°

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  45^@ = h/y`

`=> 1 = h/y`

`=> y = h`

Again in a triangle ADB

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

`=> tan 45^@ = h/y`

`=> tan 30^@ = h/(2x + y)`

`=> 1/sqrt3 = h/(2x + y)`

`=> sqrt3h = 2x + y`

`=> h(sqrt3 - 1) = 2x`

`=> h = (2x)/(sqrt3 - 1)`

`=> h = x(sqrt3 + `)`

Hence the height of tower is `x(sqrt3 + 1)` m