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
