The angle of elevation of the top of a tower standing on a horizontal plane from a point A is α. After walking a distance d towards the foot of the tower the angle of elevation is found to be β. The height of the tower is

#### Options

\[\frac{d}{cot \alpha + cot \beta}\]

\[\frac{d}{cot \alpha + cot \beta}\]

\[\frac{d}{\tan \beta - \tan \alpha}\]

\[\frac{d}{\tan \beta - \tan \alpha}\]

#### Solution

The given information can be represented with the help of a diagram as below.

Here, *CD* = *h* is the height of the tower. Length of *BC* is taken as *x*.

In`Δ ACD`

`tan A=(CD)/(AC)`

`tan∝=h/(d+x)`

`h=(d+x)tan∝ `............(1)

In ΔBCD.

`tan ß = CD/BC`

`tan ß=h/x`

`x=h cot ß` ...............(2)

From (1) and (2)

`h=(d+h cot ß)tan ∝`

`h=d tan ∝+h cot ß tan ∝`

`h(1-cot ß tan ∝ )= d tan ∝`

`h=d tan ∝/((1-cot ß tan ∝ ))=d/(cot ∝-cot ß)`