Q.The heights of two pillars are 45 meters and 15 meters respectively. From the base of the second pillar, the angle of elevation to the top of the first pillar is 60°. Determine the angle of elevation to the top of the second pillar from the base of the first pillar.

Let us assume two pillars AB = 45 meters and CD = 15 meters. From the base of the second pillar (point D), the angle of elevation to the top of the first pillar (point A) is ∠ADB = 60°. We are required to find the angle of elevation from the base of the first pillar (point B) to the top of the second pillar (point C), i.e., ∠CBD. Let ∠CBD = θ. From right-angled triangle ABD: \(\tan 60^\circ = \cfrac{AB}{BD}\) Or, \(\sqrt{3} = \cfrac{AB}{BD}\) Or, \(BD = \cfrac{AB}{\sqrt{3}}\) Or, \(BD = \cfrac{45}{\sqrt{3}} = \cfrac{45 \times \sqrt{3}}{3} = 15\sqrt{3}\) Now, from right-angled triangle CBD: \(\tan ∠CBD = \cfrac{CD}{BD}\) Or, \(\tan θ = \cfrac{15}{15\sqrt{3}}\) Or, \(\tan θ = \cfrac{1}{\sqrt{3}}\) Or, \(\tan θ = \tan 30^\circ\) Therefore, \(θ = 30^\circ\) ∴ The angle of elevation from the base of the first pillar to the top of the second pillar is \(30^\circ\).
Similar Questions