Answer :

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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