There is an airplane at an altitude of 12000 ft. The angle of depression is 1 degree. How far on the ground is the plane.
Did you make a sketch?
if so , you will see that
tan 1° = 12000/x , where x is the distance along the ground
x = 12000/tan 1° = .....
"How far on the ground is the plane ? " sounds like a very strange question.
To find the distance on the ground, we can use basic trigonometry.
Let's assume that the distance on the ground is represented by x. We have an angle of depression of 1 degree, which means that the angle formed between the line of sight from the observer to the airplane and the horizontal ground is 1 degree.
In a right triangle, the angle of depression is the angle between the hypotenuse (the line of sight) and the horizontal ground. The side opposite the angle of depression is the altitude of the airplane (12000 ft), and the side adjacent to the angle of depression is the distance on the ground (x).
Using the trigonometric function tangent (tan), we can set up the following equation:
tan(1 degree) = opposite/adjacent
tan(1 degree) = 12000/x
Now, we can solve for x by rearranging the equation:
x = 12000 / tan(1 degree)
Using a scientific calculator, we can calculate the value of x.
Note: Make sure your calculator is set to degree mode when calculating trigonometric functions in degrees.
x ≈ 685942 ft
Therefore, the airplane is approximately 685,942 ft away from the observer on the ground.