1. The top of 15 meters high tower makes an angle of depression of 60 degrees with the bottom of a electric pole and an angle of 30 degrees with the top of the pole. What is the height of the electric pole?
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By: anil on 05 May 2019 01.34 am
AD is the tower = 15 m and CE is the electric pole Let AB = $$x$$ m and DE = BC = $$y$$ m Also, $$angle$$ AED = 60° and $$angle$$ ACB = 30° In $$ riangle$$ ADE, => $$tan(angle AED)=frac{AD}{DE}$$ => $$tan(60)=sqrt{3}=frac{15}{y}$$ => $$y=frac{15}{sqrt{3}}=5sqrt{3}$$ m In $$ riangle$$ ABC, => $$tan(angle ACB)=frac{AB}{BC}$$ => $$tan(30)=frac{1}{sqrt{3}}=frac{x}{5sqrt{3}}$$ => $$x=5$$ m $$ herefore$$ CE = AD - AB = 15 - 5 = 10 meters => Ans - (C)
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