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Height and Distance
Height and Distance
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Question

The angle of elevation of a jet plane from a point A on the ground is 60∘^\circ. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30∘^\circ. If the jet plane is flying at a constant height, then its height is :

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Solution

v=432×100060×60v = 432 \times {{1000} \over {60 \times 60}} m/sec = 120 m/sec Distance AB = v ×\times 20 = 2400 meter In Δ\DeltaPAC tan⁡60∘=hPC⇒PC=h3\tan 60^\circ = {h \over {PC}} \Rightarrow PC = {h \over {\sqrt 3 }} In Δ\DeltaPBD tan⁡30∘=hPD⇒PD=3h\tan 30^\circ = {h \over {PD}} \Rightarrow PD = \sqrt 3 h PD = PC + CD 3h=h3+2400⇒2h3=2400\sqrt 3 h = {h \over {\sqrt 3 }} + 2400 \Rightarrow {{2h} \over {\sqrt 3 }} = 2400 h=12003h = 1200\sqrt 3 meter

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