There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.
In fig let the height of the other temple is h (m.) and distance between two temple is x (m.)
In ∆ABC
tan 60° =
tan 60° =
√3 =
x = ⇒
x = (1)
In ∆AED
tan 30° =
=
X = h√3 (2)
From eqn. (1) and eqn. (2) we get,
√3h =
h = = 16.67m
On substituting the value of ‘h’ in eqn (2)
X = ⇒ 28.87m
Therefore height of the temple is 50-16.67 = 33.33m
And the distance between the two temples is 28.87m