Answer: D
From triangle ABC, with respect to angle ∠ACB: \( \tan 30^\circ = \cfrac{AB}{BC} \) ⇒ \( \cfrac{1}{\sqrt{3}} = \cfrac{AB}{BC} \) ⇒ \( AB = \cfrac{BC}{\sqrt{3}} \) — (i) From triangle ABD, with respect to angle ∠ADB: \( \tan 60^\circ = \cfrac{AB}{BD} \) ⇒ \( \sqrt{3} = \cfrac{AB}{BD} \) ⇒ \( AB = \sqrt{3} \cdot BD \) — (ii) Therefore, \( \cfrac{BC}{\sqrt{3}} = \sqrt{3} \cdot BD \) ⇒ \( BC = 3 \cdot BD \) ⇒ \( BC = 3(BC - 40) \) ⇒ \( BC - 3BC = -120 \) ⇒ \( -2BC = -120 \) ⇒ \( BC = 60 \) Hence, \( AB = \cfrac{BC}{\sqrt{3}} = \cfrac{60}{\sqrt{3}} = \cfrac{60\sqrt{3}}{3} = 20\sqrt{3} \) ∴ The height of the rod is \( 20\sqrt{3} \) meters.
From triangle ABC, with respect to angle ∠ACB: \( \tan 30^\circ = \cfrac{AB}{BC} \) ⇒ \( \cfrac{1}{\sqrt{3}} = \cfrac{AB}{BC} \) ⇒ \( AB = \cfrac{BC}{\sqrt{3}} \) — (i) From triangle ABD, with respect to angle ∠ADB: \( \tan 60^\circ = \cfrac{AB}{BD} \) ⇒ \( \sqrt{3} = \cfrac{AB}{BD} \) ⇒ \( AB = \sqrt{3} \cdot BD \) — (ii) Therefore, \( \cfrac{BC}{\sqrt{3}} = \sqrt{3} \cdot BD \) ⇒ \( BC = 3 \cdot BD \) ⇒ \( BC = 3(BC - 40) \) ⇒ \( BC - 3BC = -120 \) ⇒ \( -2BC = -120 \) ⇒ \( BC = 60 \) Hence, \( AB = \cfrac{BC}{\sqrt{3}} = \cfrac{60}{\sqrt{3}} = \cfrac{60\sqrt{3}}{3} = 20\sqrt{3} \) ∴ The height of the rod is \( 20\sqrt{3} \) meters.