Question

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

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

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

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

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

Answer 1

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)`