#### Question

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