A pilot flying at an attitude of 25,000 ft. The angle of depression from her position to the start of the runway is 2 degrees. How far is the airplane from start of the runway ( in ground distance)?
Thank you
As always, draw a diagram. You will see that
x/25000 = tan2°
To find the ground distance between the airplane and the start of the runway, we can use trigonometry.
First, let's visualize the problem. We have a right triangle formed by the airplane, the point directly below it on the ground (start of the runway), and the line of sight connecting these two points.
The angle of depression is the angle between the line of sight and the horizontal plane (ground). In this case, the angle of depression is given as 2 degrees.
Next, we can use the tangent function (tan) to find the distance. The tangent of an angle is equal to the opposite side divided by the adjacent side in a right triangle.
In this case, the opposite side is the difference in altitude between the airplane and the start of the runway, which is 25,000 ft. The adjacent side is the ground distance we need to find, let's call it "x". So we have:
tan(2 degrees) = opposite/adjacent = 25,000 ft / x
Now, we can solve for "x". Rearranging the equation, we get:
x = 25,000 ft / tan(2 degrees)
Using a scientific calculator or an online trigonometry calculator, we can find the tangent of 2 degrees (tan(2)) which is approximately 0.0349. Plugging in this value, we get:
x = 25,000 ft / 0.0349
x ≈ 716,618 ft
Therefore, the airplane is approximately 716,618 ft away from the start of the runway in ground distance.