Answer: D
From triangle ABC, with respect to angle ∠ACB: \[ \tan 30^\circ = \frac{AB}{BC} \] So, \[ \frac{1}{\sqrt{3}} = \frac{AB}{BC} \Rightarrow AB = \frac{BC}{\sqrt{3}} \quad \text{---(i)} \] From triangle ABD, with respect to angle ∠ADB: \[ \tan 60^\circ = \frac{AB}{BD} \] So, \[ \sqrt{3} = \frac{AB}{BD} \Rightarrow AB = \sqrt{3} \cdot BD \quad \text{---(ii)} \] Therefore, equating (i) and (ii): \[ \frac{BC}{\sqrt{3}} = \sqrt{3} \cdot BD \Rightarrow BC = 3BD \] Now, since the shadow length increases by 40 meters: \[ BC = 3(BC - 40) \] Solving: \[ BC - 3BC = -120 \Rightarrow -2BC = -120 \Rightarrow BC = 60 \] Thus, \[ AB = \frac{BC}{\sqrt{3}} = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \] ∴ The height of the rod is \(20\sqrt{3}\) meters.
From triangle ABC, with respect to angle ∠ACB: \[ \tan 30^\circ = \frac{AB}{BC} \] So, \[ \frac{1}{\sqrt{3}} = \frac{AB}{BC} \Rightarrow AB = \frac{BC}{\sqrt{3}} \quad \text{---(i)} \] From triangle ABD, with respect to angle ∠ADB: \[ \tan 60^\circ = \frac{AB}{BD} \] So, \[ \sqrt{3} = \frac{AB}{BD} \Rightarrow AB = \sqrt{3} \cdot BD \quad \text{---(ii)} \] Therefore, equating (i) and (ii): \[ \frac{BC}{\sqrt{3}} = \sqrt{3} \cdot BD \Rightarrow BC = 3BD \] Now, since the shadow length increases by 40 meters: \[ BC = 3(BC - 40) \] Solving: \[ BC - 3BC = -120 \Rightarrow -2BC = -120 \Rightarrow BC = 60 \] Thus, \[ AB = \frac{BC}{\sqrt{3}} = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \] ∴ The height of the rod is \(20\sqrt{3}\) meters.