A plane is flying at a constant altitude and speed of 500km/hr. At a certain point, the angle of depression from the airplane to a lighthouse on an island in the distance is 6 degrees. After 12 minutes, as the plane continues to approach the lighthouse, the angle of depression becomes 15 degrees. How long will it take for the plane to be directly above the lighthouse from the point where the angle of depression was 6 degrees
in 12 minutes, the plane has gone 100km
So, reviewing your basic trig functions, you should see that at height h,
h cot 6° - h cot 15° = 100
h = 17.29km
Now the distance to the point over the lighthouse is
h tan75° = 64.53km
At 500 km/hr, that will take 7.75 minutes to cover
To solve this problem, we can use trigonometry. Let's denote the time it takes for the plane to be directly above the lighthouse as "t" (in minutes).
Step 1: Convert the angles from degrees to radians.
Angle of depression at 6 degrees = 6 * (π/180) = π/30
Angle of depression at 15 degrees = 15 * (π/180) = π/12
Step 2: Determine the distance traveled by the plane from the initial point to where the angle of depression is 15 degrees.
Since the plane is flying at a constant speed of 500 km/hr, the distance it travels in 12 minutes is (12/60) * 500 = 100 km.
Step 3: Set up a trigonometric equation using the tangent function.
tan(π/12) = h/(100 + 500t)
Step 4: Solve the equation for "t".
We need to find the time when the angle of depression is 6 degrees (π/30). So, we substitute this value into the equation and solve for "t".
tan(π/30) = h/(100 + 500t)
Using a calculator to find the tangent of π/30, we get:
0.104528 = h/(100 + 500t)
Cross-multiplying:
0.104528 * (100 + 500t) = h
Multiplying the left side of the equation:
10.4528 + 52.264t = h
Since h represents the altitude and the plane is flying at a constant altitude, the value of h remains the same at both the initial and final points.
Setting the two equations for h equal to each other:
10.4528 + 52.264t = 500t
Simplifying the equation by subtracting 52.264t from both sides:
10.4528 = 500t - 52.264t
Combining like terms:
10.4528 = 447.736t
Dividing both sides by 447.736:
t = 10.4528 / 447.736
Using a calculator to compute the division, we find:
t ≈ 0.0234
So, it will take approximately 0.0234 minutes for the plane to be directly above the lighthouse from the point where the angle of depression was 6 degrees.
To solve this problem, we can use trigonometry. Let's first understand the situation and make some observations:
1. The angle of depression is the angle between a line from the observer (plane) to the object (lighthouse) and the horizontal ground.
2. As the plane approaches the lighthouse, the angle of depression increases.
3. When the plane is directly above the lighthouse, the angle of depression will be 90 degrees.
4. We are given the initial angle of depression (6 degrees) and the time it takes for the angle of depression to increase to 15 degrees (12 minutes).
Now, let's start solving the problem:
Step 1: Find the vertical distance the plane travels from the point where the angle of depression was 6 degrees to where it becomes 15 degrees.
Since the plane is flying at a constant speed of 500 km/hr, in 12 minutes (or 12/60 = 0.2 hours), it will cover a horizontal distance of 500 * 0.2 = 100 km.
To find the vertical distance traveled, we can use the tangent function. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the opposite side represents the change in vertical distance, and the adjacent side represents the horizontal distance traveled.
Let x be the vertical distance traveled by the plane. We have:
tan(6 degrees) = x / 100 km
Solving for x, we get:
x = 100 km * tan(6 degrees)
Step 2: Find the time it takes for the plane to be directly above the lighthouse.
When the plane is directly above the lighthouse, the angle of depression is 90 degrees. Let's assume t is the time it takes for this to happen.
Using similar reasoning as in Step 1, the vertical distance traveled in t hours is:
x = 500 km/hr * t
Since the vertical distance traveled is the same as in Step 1, we can set the two equations equal to each other:
100 km * tan(6 degrees) = 500 km/hr * t
Solving for t, we get:
t = (100 km * tan(6 degrees)) / (500 km/hr)
Now, we can calculate the value of t.
Using a scientific calculator or trigonometric tables, we find that:
tan(6 degrees) ≈ 0.1051
Substituting this value into the equation, we get:
t = (100 km * 0.1051) / (500 km/hr)
Calculating this expression, we find:
t ≈ 0.02102 hours
Finally, we can convert hours to minutes by multiplying by 60:
t ≈ 1.26 minutes
Therefore, it will take approximately 1.26 minutes for the plane to be directly above the lighthouse from the point where the angle of depression was 6 degrees.