The height of a tower is 50√3 m. The angle of elevation of a tower from a distance 50 m from its feet is ?->(Show Answer!)

1. The height of a tower is 50√3 m. The angle of elevation of a tower from a distance 50 m from its feet is

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By: anil on 05 May 2019 01.49 am

Given : AB is the tower = $$50sqrt{3}$$ m and BC = 50 m To find : $$angle$$ ACB = $$ heta$$ = ? Solution : In $$ riangle$$ ABC => $$tan( heta)=frac{AB}{BC}$$ => $$tan( heta)=frac{50sqrt{3}}{50}$$ => $$tan( heta)=sqrt{3}$$ => $$ heta=tan^{-1}(sqrt{3})$$ => $$ heta=60^circ$$ => Ans - (C)

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