A plane flying at an altitude of 13,000 feet finds the angle of depression for a building ahead of the plane to be 10.1°. With the building still straight ahead, two minutes later the angle of depression is 25.8°. Find the speed of the plane relative to the ground (in feet per minute).
Make a sketch showing the two positions of the plane.
Let the speed of the plane be x ft/min
Let the horizontal distance of the plane from the building be d
(along the ground)
tan 10.1° = 13000/d, d = 13000/tan10.1 = 72,981.6 ft
let the horizontal distance of the plane from the building be k for the second position,
tan 25.8° = 13000/k, k = 13000/tan25.8 = 26,891.8 ft
difference = d - k = 46,089.8 ft
rate = distance/time = 46,089.8/2 ft/min = 23,044.9 ft/min
(which would be about 262 mph, making sense)
Check my arithmetic
To find the speed of the plane relative to the ground, we can use the tangent formula:
tan(angle of depression) = height of building / distance from the plane to the building
Let's denote the distance from the plane to the building as 'x'.
From the first observation, we have:
tan(10.1°) = height of building / x
Simplifying, we get:
x = height of building / tan(10.1°)
Similarly, from the second observation, we have:
tan(25.8°) = height of building / (x + 2 minutes * speed of the plane)
We need to convert the 2 minutes to feet, since the height of the building and the distance are given in feet.
1 minute = 60 seconds
1 mile = 5280 feet
1 hour = 60 minutes
So, 2 minutes = 2 * 60 = 120 seconds
The speed of the plane = x feet / 120 seconds
Substituting these values into the second equation, we get:
tan(25.8°) = height of building / (x + (x/120) * 120)
Simplifying, we get:
tan(25.8°) = height of building / (2x)
Now we have two equations:
x = height of building / tan(10.1°)
tan(25.8°) = height of building / (2x)
We can solve these two equations simultaneously to find the speed of the plane relative to the ground.
To find the speed of the plane relative to the ground, we need to determine how much the plane has traveled horizontally in those two minutes.
First, let's label the known quantities:
- Altitude of the plane, h = 13,000 feet
- Initial angle of depression, θ1 = 10.1°
- Final angle of depression after 2 minutes, θ2 = 25.8°
We can use trigonometry, specifically the tangent function, to relate the angles of depression and the horizontal distance traveled by the plane.
Step 1: Find the horizontal distance, x1, when the initial angle of depression θ1 = 10.1°.
tan(θ1) = h / x1 (opposite/adjacent side)
x1 = h / tan(θ1)
Step 2: Find the horizontal distance, x2, when the final angle of depression θ2 = 25.8°.
x2 = h / tan(θ2)
Step 3: Calculate the change in horizontal distance in 2 minutes.
Δx = x2 - x1
Step 4: Determine the speed of the plane relative to the ground in feet per minute.
Speed = Δx / 2 (since the change occurred over 2 minutes)
Now let's plug in the values and solve for the speed:
x1 = 13,000 / tan(10.1°)
x2 = 13,000 / tan(25.8°)
Δx = x2 - x1
Speed = Δx / 2
Using a calculator to evaluate the tangent values and perform the calculations, we can find the speed of the plane relative to the ground in feet per minute.