Q. 204.1( 9 Votes )

An observer

Answer :

20.jpg


Let height of the tower CD be h m.


Consider ΔDCB,


tan45° = DC/CB


1 = h/CB


CB = h


Consider ΔDCA,


tan30° = DC/CA


1/√3 = h/CA


Since CA = CB + AB = h + 20


1/√3 = h/h + 20


h + 20 = √3h


√3h – h = 20


h (√3 – 1) = 20


h = 20/ (√3 – 1)


h = 10 + 10√3


h = 27.32 m


Ans. The height of the tower is 27.32 m.

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