Q.The heights of two pillars are 180 meters and 60 meters respectively. If the angle of elevation from the base of the second pillar to the top of the first pillar is 60°, find the angle of elevation from the base of the first pillar to the top of the second pillar.

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 \(30°\).
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