Triangle Trig- (Pre-cal)?

Question

I need help with these problems:

1. An airplane is at an elevation of 35,000 ft when it begins its approach to an airport. Its angle of descent is 6 degrees.
(a) what is the distance between the airport and the point on the ground directly below the airplane?
(b) What is the approximate air distance between the plane and the airport?

Answer 1

If the plane is descending at a 6 degree angle, then it must be tilted at 84 degrees from vertical.

Then tan(84) = horizontal distance to airport / 35000.

Solving for horizontal distance to airport = 35000(tan 84) = 333,002 feet.

Using the Pythagorean theorem, we can compute the distance between the plane and the airport as Sqrt((333,002)^2 + (35000)^2) = 334,836 feet.

Answer 2

a) 35,000/d = tan 6°
d = 35,000/tan 6° = 333002.8 ft = 63.07 mi

b) 35,000/a = sin 6°
a = 35,000/sin 6 = 334837.0 ft = 63.42 mi.