The angles of depression of two ships from the top of a light house 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 light house.
In the fig let AB is the light house of height h (m)
In ∆ABC
tan 30° =
=
√3h = 200 + x ………………….(1)
In ∆ABD
tan 45° =
1 =
h = x ………………(2)
From eqn (1) and (2) we get
√3h = 200 + h
h(√3 + 1) = 200
h =
On multiplying and dividing by √3-1, we get
h =
h = ⇒ 273.2m
Therefore height of the light house is 273.2m