The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30° respectively. If the ships are 200 m apart, find the height of the lighthouse.
Solution
Let CD be the light house and A and B be the positions of the two ships.
AB = 200 m (Given)
Suppose CD = h m and BC = x m
Now,
∠DAC = ∠ADE = 30º (Alternate angles)
∠DBC =∠EDB = 45º (Alternate angles)
In right ∆BCD,
`tan 45^@ = (CD)/(BC)`
`=> 1 = h/x`
=> x = h .....(1)
In right ∆ACD,
`tan 30^@ = (CD)/(AC)`
`=> 1/sqrt3 = h/(x + 200)`
` => sqrt3h = x + 200` ......(2)
From (1) and (2), we get
`sqrt3h = 200 + h`
`=> sqrt3h - h = 200`
`=> (sqrt3 - 1)h = 200``
`=> h = 200/(sqrt3 - 1)`
`=> h = (200(sqrt3 + 1))/(((sqrt3 -1)(sqrt3 + 1))`
`=> h = (200(sqrt3 + 1))/2 = 100(sqrt3 + 1)`m
Hence, the height of the light house is `100(sqrt3 + 1)` m
