one hundred feet directly overhead, a plane and a dirigible start flying due east. If the plane travels three times faster than the dirigible, what is the greatest angle of sight between the two vehicles from your position? Assume the plane passes from sight after 6000 feet.
An airplane was preparing for landing.When it began its descent towards the airport, the plane's altitude was 22,000 feet.Twelve minutes later,its altitude was 6000 feet
To find the greatest angle of sight between the two vehicles, we need to determine the positions of the plane and dirigible at the point when the plane passes out of sight.
Let's first calculate the distance covered by the dirigible:
Distance covered by the dirigible = Speed × Time
Since the dirigible is traveling due east at a certain speed, its distance covered will be the same as the distance the plane travels.
Let's denote the speed of the dirigible as "S" and the time it takes for the plane to pass out of sight as "T". Therefore, the distance covered by the dirigible will be 3 times the distance covered by the plane:
Distance covered by the dirigible = 3 × Distance covered by the plane
Now, let's calculate the distance covered by the plane:
Distance covered by the plane = Speed × Time = 3S × T
We know that the plane passes out of sight after covering a distance of 6000 feet:
Distance covered by the plane = 6000 feet
Therefore, we can write the equation:
3S × T = 6000
Now, let's determine the position of the plane when it passes out of sight:
Height = 100 feet
Using the Pythagorean theorem, we can find the distance between the plane and us when it passes out of sight:
Distance^2 = (Height)^2 + (Distance covered by the plane)^2
Plugging in the values:
Distance^2 = (100)^2 + (6000)^2
Now, let's solve for the distance:
Distance = sqrt((100)^2 + (6000)^2)
Finally, let's calculate the angle of sight:
Angle of sight = arctan(Height / Distance)
Angle of sight = arctan(100 / sqrt((100)^2 + (6000)^2))
After evaluating this expression, we find that the greatest angle of sight between the two vehicles from our position is approximately 0.959 degrees.
To find the greatest angle of sight between the plane and the dirigible, we need to first determine their positions relative to you.
Let's set up a coordinate system with you as the origin. The plane is flying due east, so its x-coordinate is increasing. The dirigible is also flying due east, but at a slower speed. Since they both start at the same point, we can represent their positions using the equation x = v*t, where x represents the distance in feet, v represents the velocity in feet per second, and t represents time in seconds.
Let's denote the velocity of the dirigible as d ft/s and the velocity of the plane as 3d ft/s.
The plane passes from sight after traveling 6000 feet. Setting up an equation for the plane's position, we have:
6000 = (3d) * t
Simplifying, we find:
t = 6000 / (3d) = 2000 / d
Now, let's consider the dirigible. It maintains a slower speed, so its position can be represented by:
x = d * t = d * (2000 / d) = 2000
The dirigible reaches a maximum distance of 2000 feet directly east from you.
To calculate the greatest angle of sight, we can use trigonometry. The angle can be found using the tangent function, which is equal to the opposite side (height) divided by the adjacent side (distance):
tan(theta) = height / distance
In this case, the height is 100 feet (given in the question), and the distance is 2000 feet.
tan(theta) = 100 / 2000 = 1 / 20
Now, we can find the angle by taking the arctan (inverse tangent) of both sides:
theta = arctan(1/20) ≈ 2.86 degrees
Therefore, the greatest angle of sight between the plane and the dirigible from your position is approximately 2.86 degrees.