Answer :

In the fig AB is the light house of height of 200m.


In ∆AED


tan 45° =


1 =


x = 200m ………………….(1)


In ∆ABC


tan 60° =


√3 =


√3y = 200


y =


On multiplying and dividing by √3, we get


y = m


Therefore the distance between two ships is:



DC = x + y


x + y = 200 +



= 315.47m


Therefore the distance between two ships is 315.47m


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