  Question

The pilot of an aircraft flying horizontally at a speed 1200km/hr.Observes  that the angle of depression of a point on the ground c hanges from $${30^ \circ }$$ to $${45^ \circ }$$  in 15 sec.Find the height at which the aircraft is flying.

Solution

REF.Image.Solution:-Speed of an aircraft  $$= 1200 km/hr$$After 15 secs, angle of depression changes from $$30^{\circ}$$ to $$45^{\circ}$$Distance covered in 15 seconds $$= 1200 \times 15 \times 3600 = 5 km=5000m$$ Let height from ground be $$h$$Horizontal distance $$= x$$ In $$\triangle BDP, \angle P = 45^{\circ}, \therefore tan45^{\circ} = \dfrac{BD}{DP} = \dfrac{h}{DP}$$$$\therefore DP = h$$In $$\triangle ACP, \angle APC = 30^{\circ}$$$$\therefore tan30^{\circ} = \frac{AC}{CP}$$$$\therefore \dfrac{1}{\sqrt{3}} = \dfrac{h}{5000+h}, 5000+h = h\sqrt{3}$$$$\Rightarrow 5000 = h (\sqrt{3}-1)$$$$\Rightarrow h = 6830m = 6.830 km$$ (height of an aircrft from ground) Mathematics

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