There is a 12 m tall hoarding pole on the top of a building. A person at some distance from the building, observes that the angle of elevation to the top of the hoarding pole from that point is the same as the angle of elevation to the bottom of the hoarding pole if he moves 4 √ 3 m closer to the building. What is the angle of elevation that the person is seeing?
Let AB be the building of height ‘h’ and BC be the hoarding pole of height 12 m. Let θ be the angle of elevation as seen by the person at a distance of ‘x’ m and ‘x + 4 √ 3‘ m. In Δ DAB, tan θ = h/x (i) In Δ EAC, tan θ = (h + 12)/ (x + 4 √ 3) (ii) From (i) and (ii) h/x = (h + 12)/ (x + 4 √ 3) ⇒ hx + 4 √ 3h = hx +12x ⇒ h/x = 12/ 4 √ 3 = √ 3 ⇒ tan θ = h/x = √ 3 = tan 60° ⇒ θ = 60°
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