A man standing on a level plane observes the elevation of the top of a pole to be θ. If he walks a distance equal to double the height of the pole towards the pole, the angle of elevation becomes 2θ. Then the value of θ (in degrees) is
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Solution
Let AB be the pole with height h.
In △ABC x=hcot2θ⋯(i)
In △ABD h=(2h+x)tanθ⋯(ii)
From equation (i) and (ii), h=(2h+hcot2θ)tanθ⇒1=(2+cot2θ)tanθ(∵h≠0)⇒1=(2+1tan2θ)tanθ⇒1=(2+1−tan2θ2tanθ)tanθ⇒tan2θ−4tanθ+1=0⇒tanθ=4±√122⇒tanθ=2+√3,2−√3⇒θ=75∘,15∘
As 2θ is an acute angle, so θ=15∘