Q. 64

# 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

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