Let AB be the height of a house. From point A on the roof, the angles of depression to the top (P) and base (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 the right-angled triangle ∆ABQ: \[ \tan 60^\circ = \frac{AB}{BQ} \Rightarrow \sqrt{3} = \frac{AB}{BQ} \Rightarrow BQ = \frac{AB}{\sqrt{3}} \] From the right-angled triangle ∆ACP: \[ \tan 30^\circ = \frac{AC}{CP} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AC}{CP} \Rightarrow CP = AC \sqrt{3} \] Since \(BQ = CP\), \[ \frac{AB}{\sqrt{3}} = AC \sqrt{3} \Rightarrow AB = 3AC \Rightarrow AB = 3(AB - BC) \Rightarrow AB = 3(AB - PQ) \quad [\text{since } BC = PQ] \Rightarrow AB = 3AB - 3PQ \Rightarrow AB - 3AB = -3PQ \Rightarrow -2AB = -3PQ \Rightarrow \frac{AB}{PQ} = \frac{3}{2} \] ∴ The ratio of the height of the house to the height of the lamp post is 3:2.