Let AB = taller pole CD = shorter pole = h OB = OD =( 12)/2 = 6m In ∆OCD, tanθ = h/6 ............................(i) In ∆AOB tan (90- θ) = 2h/6 = h/3 cotθ= h/3 .............................(ii) Multiplying (i) and (ii), we get tanθ × cotθ = h/6 × 4/3 1 = h²/18 h² = 18 h=3√2 m
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