Let AB be a 30-meter high pole and CD be a pillar. From the top of the pole (point A), the foot of the pillar (point D) is seen at a depression angle of 30°, and the top of the pillar (point C) is seen from the base of the pole (point B) at an elevation angle of 60°. A line drawn from point A parallel to BD intersects CD at point E. Since AE || BD, ∠ADB = alternate angle ∠EAD = 30° From the right-angled triangle ABD: \[ \tan 30^\circ = \frac{AB}{BD} = \frac{30}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{BD} \Rightarrow BD = 30\sqrt{3} \] Also, AE = BD = \(30\sqrt{3}\) From the right-angled triangle AEC: \[ \tan 60^\circ = \frac{CE}{AE} = \frac{CE}{30\sqrt{3}} \Rightarrow \sqrt{3} = \frac{CE}{30\sqrt{3}} \Rightarrow CE = 90 \] Now, \[ CD = CE + ED = CE + AB = 90 + 30 = 120 \] Therefore, the height of the pillar CD is 120 meters and the distance between the pole and the pillar BD is \(30\sqrt{3}\) meters.