Question

the angle of depression from the top and bottom of a tower as seen from the top of a 60√3m high cliff is 45° and 60° respectively. find the HT of the tower.
pls draw the figure also pls.....​

Answer 1

here ur answer in this attachmnt

hope u understand

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Answer 2

Refer the attachment for answer.

Let AB = h ( Height of the tower )

Thus, DE = h

Given that :

CD = 60√3  metres (Height of Cliff )

Therefore,

CE = 60√3 - h

In right angled ∆ AEC,

tan 45° = CE/EA

1 = (60√3 - h) / EA

EA = 60√3 - h...(1)

In right angled ∆ CDB,

tan 60° = CD/DB

√3 = 60√3/DB

DB = 60 metres

From (1),

60 = 60√3 - h

h = 60√3 - 60

h = 60(√3 - 1)

h = 60 ( 1.73 - 1 )

h = 60 × 0.73

h = 43.2 metres

Hence, Height of the tower is 43.2 metres.