The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.
In the fig AB is the Rock and CD is the tower.
AB = AE + EB ⇒ h + 100
In ∆ABC
tan 45° =
1 =
x = 100 + h ……………….(1)
In ∆AEB
tan 30° =
=
√3h= 100 + h
h(√3-1) = 100
h =
On multiplying and dividing by √3 + 1, we get
h = ⇒
⇒ 50 (√3 + 1) = 136.6m
Therefore height of the rock = h +100 ⇒ 136.6 + 100 = 236.6m