Let’s assume two pillars: AB = 180 meters and CD = 60 meters. From the base of the first pillar (point B), the angle of elevation to the top of the second pillar (point C) is ∠CBD = 30°. We need to find the angle of elevation ∠ADB from the base of the second pillar (point D) to the top of the first pillar (point A). Let ∠ADB = θ. From the right-angled triangle CBD: \[ \tan 30^\circ = \frac{CD}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{CD}{BD} \Rightarrow BD = CD \cdot \sqrt{3} \Rightarrow BD = 60\sqrt{3} \] From the right-angled triangle ABD: \[ \tan ∠ADB = \frac{AB}{BD} \Rightarrow \tan θ = \frac{180}{60\sqrt{3}} \Rightarrow \tan θ = \sqrt{3} \Rightarrow \tan θ = \tan 60^\circ \Rightarrow θ = 60^\circ \] ∴ The angle of elevation from the base of the second pillar to the top of the first pillar is 60°