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To approach the runway, a pilot of a small plane must begin a 10Β° descent starting from a height of 1983 feet 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?

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The plane must be 3.38 miles away from the runway to begin the descent.Β  The equation I used was "y = -0.1111111x + 0.3755682", 0.37~ being the height in feet (in miles) and -0.11~ being the descent.Β  The x-intercept is located at 3.38.
sqdancefan
The ratio of vertical distance to horizontal distance is the tangent of the angle.
Β  tan(10Β°) = (1983 ft)/(horizontal distance in ft)
Then
Β  (horizontal distance in ft) = (1983 ft)/(tan(10Β°)) β‰ˆ 11,246.15 ft
The horizontal distance in miles is found using the appropriate conversion.
Β  (horizontal distance in mi) = (horizontal distance in ft)*(1 mi)/(5280 ft)
Β  = 11,246.15/5280 mi β‰ˆ 2.129953... mi

The airplane is approximately 2.1 miles from the end of the runway.

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