Q. 43

# 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

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