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°.