Let AB be a building of height 60 meters. From the top point A, the angles of depression to the top (P) and bottom (Q) of a lamp post PQ are 30° and 60°, respectively. ∴ ∠DAP = 30° and ∠DAQ = 60° [Assume AD ∥ BQ] Draw CP ∥ BQ. Then, ∠APC = alternate angle ∠DAP = 30° ∠AQB = alternate angle ∠DAQ = 60° From right-angled triangle ∆ABQ: \[ \tan 60^\circ = \frac{AB}{BQ} \] \[ \Rightarrow \sqrt{3} = \frac{AB}{BQ} \] \[ \Rightarrow BQ = \frac{AB}{\sqrt{3}} = \frac{60}{\sqrt{3}} = \frac{60\sqrt{3}}{3} = 20\sqrt{3} \] ∴ The horizontal distance between the building and the lamp post is \(20\sqrt{3}\) meters. **[Answer to part (i)]** Now, from right-angled triangle ∆ACP: \[ \tan 30^\circ = \frac{AC}{CP} \] \[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{AC}{CP} \] \[ \Rightarrow AC = \frac{CP}{\sqrt{3}} = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \] ∴ The height of the lamp post is \[ PQ = BC = AB - AC = 60 - 20 = 40 \text{ meters} \] [Answer to part (ii)]