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Height and Distance
Height and Distance
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Question

An aeroplane flying at a constant speed, parallel to the horizontal ground, 3\sqrt 3 kmabove it, is obsered at an elevation of 60o{60^o} from a point on the ground. If, after five seconds, its elevation from the same point, is 30o{30^o}, then the speed (in km / hr) of the aeroplane, is :

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Solution

For Δ\Delta OA, A, OA 1 = 3tan⁡60o{{\sqrt 3 } \over {\tan {{60}^o}}} = 1 km For Δ\Delta OB 1 , B, OB 1 = 3tan⁡30o{{\sqrt 3 } \over {\tan {{30}^o}}} = 3km As, a distance of 3 −- 1 = 2 km is convered in 5 seconds. Therefore the speed of the plane is 2×36005{{2 \times 3600} \over 5} = 1440 km/hr

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