Answer :

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


In ∆ABC


tan 30° =


=


x = 100 - √3 h ……………(1)


In ∆ABD


tan 45° =


1=


x = h …………(2)


On substituting value of x from eqn (2) in eqn (1)



h = 100 - √3 h


h + √3 h = 100


h (1 + √3) = 100


h = = 50 (√3 - 1)


Therefore height of the light house is 50 (√3 - 1)m


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