Math Problem Statement

Samantha spots an airplane on radar that is currently approaching in a straight line, and that will fly directly overhead. The plane maintains a constant altitude of 63756375 feet. Samantha initially measures an angle of elevation of 15, degrees15 ∘ to the plane at point AA. At some later time, she measures an angle of elevation of 28, degrees28 ∘ to the plane at point BB. Find the distance the plane traveled from point AA to point BB. Round your answer to the nearest tenth of a foot if necessary.

Solution

We can approach this problem by using trigonometry, specifically the tangent function, which relates the angle of elevation, the height of the plane, and the horizontal distance between the observer (Samantha) and the plane.

Step 1: Set up the scenario

We know the following information:

  • Height of the plane: h=63756375feeth = 63756375 \, \text{feet}
  • Angle of elevation at point A: θ1=15\theta_1 = 15^\circ
  • Angle of elevation at point B: θ2=28\theta_2 = 28^\circ

Let:

  • d1d_1 be the horizontal distance from point A to the point directly below the plane.
  • d2d_2 be the horizontal distance from point B to the point directly below the plane.

We are tasked with finding the distance the plane has traveled between point A and point B, which is simply the difference in horizontal distances: distance=d2d1\text{distance} = d_2 - d_1.

Step 2: Use the tangent function

The tangent of an angle in a right triangle is defined as the ratio of the opposite side (the height of the plane) to the adjacent side (the horizontal distance to the point directly beneath the plane). So, we can write the following equations:

  1. tan(θ1)=hd1\tan(\theta_1) = \frac{h}{d_1} tan(15)=63756375d1\tan(15^\circ) = \frac{63756375}{d_1} Solving for d1d_1: d1=63756375tan(15)d_1 = \frac{63756375}{\tan(15^\circ)}

  2. tan(θ2)=hd2\tan(\theta_2) = \frac{h}{d_2} tan(28)=63756375d2\tan(28^\circ) = \frac{63756375}{d_2} Solving for d2d_2: d2=63756375tan(28)d_2 = \frac{63756375}{\tan(28^\circ)}

Step 3: Calculate the distances

Now, let's calculate the values for d1d_1 and d2d_2 using the given angles.

First, compute the values for the tangents of 1515^\circ and 2828^\circ:

  • tan(15)0.2679\tan(15^\circ) \approx 0.2679
  • tan(28)0.5317\tan(28^\circ) \approx 0.5317

So:

  • d1=637563750.2679237106524.8feetd_1 = \frac{63756375}{0.2679} \approx 237106524.8 \, \text{feet}
  • d2=637563750.5317119866671.4feetd_2 = \frac{63756375}{0.5317} \approx 119866671.4 \, \text{feet}

Step 4: Find the distance the plane traveled

The distance traveled by the plane from point A to point B is the difference between d1d_1 and d2d_2:

Distance=d1d2=237106524.8119866671.4117239853.4feet\text{Distance} = d_1 - d_2 = 237106524.8 - 119866671.4 \approx 117239853.4 \, \text{feet}

Final Answer:

The plane traveled approximately 117,239,853.4 feet from point A to point B.


Would you like any further clarifications or details on the steps? Here are some related questions for you:

  1. How do you apply the tangent function in other real-world problems involving elevation angles?
  2. What are some other methods to solve this problem if you don't have a radar and just use physical measurements?
  3. How would this problem change if the plane was at an angle instead of directly overhead?
  4. What is the relationship between the angle of elevation and the horizontal distance in real-world navigation systems?
  5. How could you estimate the time it took for the plane to travel this distance if you knew its speed?

Tip: Make sure to always convert angles to the correct units (degrees or radians) depending on the calculator or method you are using!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent Function
Right Triangles

Formulas

tan(θ) = opposite/adjacent

Theorems

Tangent theorem for right triangles

Suitable Grade Level

Grades 9-11