Two boats approach a lighthouse in mid-sea from opposite directions. The angles of elevation of the top of the lighthouse from two boats are 30° and 45° respectively. If the distance between two boats is 100 m, find the height of the lighthouse.

#### Solution

Let h be the height of lighthouse BD. An angle of elevation of the top of the lighthouse from two boats is 30° and 45°.

Let DB = h , BC = x and it is given that AC = 100 m. So

AB = 100 - x. And ∠DAB = 30°, ∠BCD = 45°

Here we have to find height of light house.

The corresponding figure is as follows

So we use trigonometric ratios.

In ΔBDC

`=> tan 45° = (BD)/(BC)`

`=> 1 = h/x`

`=> x = h `

Again in ΔDAB

`=> tan 30° = (DB)/(AB)`

`=> 1/sqrt3 = h/(100 - x)`

`=> sqrt3h = 100 - x`

`=> (sqrt3 + 1)h = 100`

`=> h = 100/(sqrt3 + 1) xx (sqrt3 - 1)/(sqrt3 - 1)`

`=> h = 50(sqrt3 - 1)`

Hence the height of light house is `50(sqrt3 - 1)` m