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  • The angle of elevation of an airplane from point A on the ground is 60°. After a flight of 10 seconds, on the same height, the angle of elevation from point A becomes 30°. If the airplane is flying at the speed of  720 km/hr, find the cons

The angle of elevation of an airplane from point A on the ground is 60°. After a flight of 10 seconds, on the same height, the angle of elevation from point A becomes 30°. If the airplane is flying at the speed of  720 km/hr, find the constant height at which the airplane is flying. 

 

Answers (1)

Given, \angle BAE =60^{\circ}

            \angle CAD=30^{\circ}

Assume BE = CD = h meter

Speed of the airplane = 720km/hr

                                    = 720x5/18 = 200 m/sec

Time = 10 seconds

Distance covered (BC) = 200x10 = 2000 meter = ED

\\$In triangle ABE$ \\ \tan 60^0 = \frac{BE}{AE} \\ \sqrt{3} = \frac{BE}{AE} \\\\ AE = \frac{h}{\sqrt{3}} \\

\\$In triangle ACD$ \\ \tan 30^0 = \frac{CD}{AD} \\\\ \frac{1}{\sqrt{3}} = \frac{h}{AD} \\\\ AD = h\sqrt{3} \\?

\\ED = AD - AE \\ 2000 = h\sqrt{3} - \frac{h}{\sqrt{3}} \\ 2000 = h(\sqrt{3} - \frac{1}{\sqrt{3}}) \\ 2000 = h( \frac{2}{\sqrt{3}}) \\ h = 1000 \sqrt{3 } \ \ m

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