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Let the height of the tower be h And ∠CBD = θ and ∠ DAC = 90 – θ In ∆ BCD tan θ = CD/BC = h/9 …………(i) In ∆ ACD tan(90 - θ) = CD/AC cot θ = h/16 ……………………….(ii) On multiplying Equation (i) and (ii) tan θ × cot θ = h/9 × h/16 1 = h²/144 Now h = 12 m
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