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