Answer :

In the fig let AB is the light house of height h (m)


In ∆ABC



tan 30° =


=


√3h = 200 + x ………………….(1)


In ∆ABD


tan 45° =


1 =


h = x ………………(2)


From eqn (1) and (2) we get


√3h = 200 + h


h(√3 + 1) = 200


h =


On multiplying and dividing by √3-1, we get


h =


h = 273.2m


Therefore height of the light house is 273.2m


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