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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)

1099834_1166582_ans_46d5565c6eeb4663a2209a0d6207ee43.jpg

Mathematics

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