Assume Habu is standing at point A on field BC and observes the bird at point P initially, and after 2 minutes, at point Q. Since the bird is flying at a constant height, PQ is parallel to BC, and the height of the bird from point A is AM = \(50\sqrt{3}\) meters. Now, ∠APM = corresponding angle ∠BAP = 30° and ∠AQM = corresponding angle ∠QAC = 60° [∵ PQ ∥ BC] From right triangle APM: \[ \tan 30^\circ = \cfrac{AM}{PM} = \cfrac{50\sqrt{3}}{PM} \Rightarrow \cfrac{1}{\sqrt{3}} = \cfrac{50\sqrt{3}}{PM} \Rightarrow PM = 150 \] From right triangle AMQ: \[ \tan 60^\circ = \cfrac{AM}{MQ} \Rightarrow \sqrt{3} = \cfrac{50\sqrt{3}}{MQ} \Rightarrow MQ = 50 \] ∴ \(PQ = PM + MQ = 150 + 50 = 200\) meters So, the bird covers 200 meters in 2 minutes. ∴ Bird's speed = \(\cfrac{200 \times 60}{2} = 6000\) meters/hour = 6 kilometers per hour