To approach the runway, a small plane must begin a 9 degree descent starting from a height of 2 miles above the ground. To the nearest tenth of a mile, how many miles from the runway is the airplane at the start of this approach ?

Question

anonymous · Answer

Since the plane has a 9 degree descent, the two sides of the triangle in question are 1960/5280 = 0.37121 mi and k miles (we don't know k yet). Then angle in question we need is actually 81 degrees--the 9 degree descent is *outside* the triangle in question. tan(81) = k/0.37121 --> k = 0.37121 * tan(81) = 0.37121 * 6.37375 = 2.344 miles The plane is 2.3 miles from the runway.\ http://answers.yahoo.com/question/index?qid=20090319111233AABRboF

anonymous · Answer

2/tan81'=x|dw:1342682782474:dw|

anonymous · Answer

Thank you guys very much. But my choices are...
A. 7mi
B. 41 mi
C. 28 mi
D. 16mi

anonymous · Answer

$$Tangent of an \angle = side opposite / side adjacent $$

$$Tangent (9\deg) = 1,960 feet / side adjacent $$

$$side adjacent = 1,960 feet / \tan(9 \deg) $$

$$Side adjacent = 12,374.95 feet $$
You want your distance in miles, though. Use the fact that 1 mile = 5,280 feet.

$$Side adjacent = (12,374.95 feet) * (1\ mile / 5280 \ feet) $$

Side adjacent = 2.34 miles. Round to the nearest tenth to get:

Side adjacent = 2.3 miles.