Q.Mohit observed a flying bird first at an elevation angle of 30° to the north, and then at an elevation angle of 60° to the south. If the bird was flying along a straight path at a height of \(50\sqrt{3}\) meters, calculate its speed in kilometers per hour.

Let’s assume Mohit is standing at point A on the field BC and observes the bird first at point P and then, 2 minutes later, 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 = alternate angle ∠BAP = 30° and ∠AQM = alternate angle ∠QAC = 60° [Since PQ || BC] From right-angled triangle APM: \(\tan 30^\circ = \frac{AM}{PM} = \frac{50\sqrt{3}}{PM}\) ⇒ \(\frac{1}{\sqrt{3}} = \frac{50\sqrt{3}}{PM}\) ⇒ \(PM = 150\) From right-angled triangle AMQ: \(\tan 60^\circ = \frac{AM}{MQ}\) ⇒ \(\sqrt{3} = \frac{50\sqrt{3}}{MQ}\) ⇒ \(MQ = \frac{50\sqrt{3}}{\sqrt{3}} = 50\) ∴ \(PQ = PM + MQ = 150 + 50 = 200\) meters So, the bird travels 200 meters in 2 minutes. ∴ Speed of the bird per hour = \(\frac{200 × 60}{2} = 6000\) meters = 6 kilometers per hour.
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