An airplane is at an elevation of 35,000 ft when it its approach to an airport. Its angle of descent is 6°.

1 answer:

7 0

Let the ground distance be x
tan6=35000÷x
x=333000 ft (corr. to 3 sig. fig.)

Let the air distance be y
sin6=35000÷y
y=335000 ft (corr. to 3 sig.fig.)

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