Q.The ratio of the heights of two pillars is 1:3. If the angle of elevation from the base of the shorter pillar to the top of the taller pillar is 60°, then find the angle of elevation from the base of the taller pillar to the top of the shorter pillar.

Let the height of the shorter pillar CD be \(h\) units. ∴ The height of the taller pillar AB is \(3h\) units. Now, from the base point D of the shorter pillar, the angle of elevation to the top A of the taller pillar is \(\angle ADB = 60^\circ\). We need to find the angle of elevation \(\angle CBD = \theta\) from the base point B of the taller pillar to the top point C of the shorter pillar. From the right-angled triangle ABD: \[ \tan 60^\circ = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{AB}{BD} \Rightarrow BD = \frac{AB}{\sqrt{3}} = \frac{3h}{\sqrt{3}} = \sqrt{3}h \] From the right-angled triangle CBD: \[ \tan \angle CBD = \frac{CD}{BD} \Rightarrow \tan \theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}} \Rightarrow \tan \theta = \tan 30^\circ \Rightarrow \theta = 30^\circ \] ∴ The angle of elevation from the base of the taller pillar to the top of the shorter pillar is 30°.
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