Answer:
The distance between the plane and the base of the airport = 103339.14 ft
Step-by-step explanation:
Consider the diagram attached.
We can notice that the a triangle (ABC) can be formed between the plane, the airport, and a vertical line drawn from the ground to the plane showing its altitude.
The angle of depression is 14 degrees. This means that angle BAC is 90 - 14 = 76 degrees.
From triangle laws, we will have that [tex]cos 76 = \frac{Altitude}{AC}[/tex]
[tex]AC = \frac{25000}{(cos 76)} = 103339.14 ft[/tex]
Hence, the distance between the plane and the base of the airport = 103339.14 ft
The diagonal distance between the plane and the base of the airport is 103339.14 ft.
This situation forms a right angle triangle.
The plane is flying at an elevation of 25000 ft.
The elevation is the height or opposite side of the triangle.
The diagonal distance between the plane and the base of the airport is the hypotenuse of the triangle.
Therefore, using trigonometric ratio,
sin 14° = opposite / hypotenuse
sin 14° = 25000 / d
d = 25000 / sin 14
d = 25000 / 0.2419218956
d = 103339.137361 ft
Therefore, the diagonal between the plane and the base of the airport is 103339.14 ft.
learn more on depression here: https://brainly.com/question/13545523?referrer=searchResults
