An aircraft takes off and climbs at a constant 29 degree angle until it reaches and altitude of 6 miles. At that point, how far away is the plane from the airport?
To determine the distance of the aircraft from the airport when it reaches an altitude of 6 miles, we can use trigonometry.
The height (opposite side) of the triangle can be represented by the altitude of 6 miles. Let's convert that to feet for consistency: 1 mile = 5280 feet, so 6 miles is equal to 6 * 5280 = 31,680 feet.
The angle of climb is given as 29 degrees, and the angle forms the hypotenuse (slant distance) of the triangle.
Now, we need to find the base (adjacent side) of the triangle, which represents the distance we're looking for.
Using trigonometry, we can use the tangent function:
tan(angle) = opposite/adjacent
In this case, tan(29 degrees) = 31,680 feet / adjacent
We can rearrange the equation to solve for the adjacent side (distance):
adjacent = opposite / tan(angle)
adjacent = 31,680 feet / tan(29 degrees)
Let's calculate the adjacent side using this formula:
adjacent = 31680 / tan(29)
Using a calculator, we find that tan(29 degrees) ≈ 0.5543.
So, adjacent = 31680 / 0.5543 ≈ 57,166.04 feet.
Therefore, when the aircraft reaches an altitude of 6 miles, it is approximately 57,166.04 feet away from the airport.
Did you make a sketch?
What you have is a right-angled triangle with a base angle of 29°, a height of 6 and the hypotenuse is what we want, call it h
6/h = sin 29°
h = 6/sin29°
= 12.38 miles