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 an altitude of 6375 feet. Samantha initially measures an angle of elevation of 15 degrees ∘ to the plane at point AA. At some later time, she measures an angle of elevation of 28 degrees ∘ 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

Let's break this problem down using trigonometry. The scenario involves two points, AA and BB, where Samantha measures the angles of elevation to an airplane at two different times. We are asked to find the distance the plane traveled between these points, given the altitude of the plane remains constant at 6375 feet.

Step 1: Visualizing the situation

We have two right-angled triangles:

  • At point AA, the angle of elevation is 1515^\circ.
  • At point BB, the angle of elevation is 2828^\circ.
  • The altitude of the plane is 63756375 feet (constant for both points).

The key is to use the tangent function for each of the angles, which relates the angle of elevation to the vertical height and horizontal distance. Specifically: tan(θ)=heighthorizontal distance\tan(\theta) = \frac{\text{height}}{\text{horizontal distance}} Let the horizontal distances from points AA and BB to the point directly beneath the plane (i.e., where the plane will pass overhead) be xAx_A and xBx_B, respectively.

Step 2: Setting up the equations

At point AA: tan(15)=6375xA\tan(15^\circ) = \frac{6375}{x_A} So, xA=6375tan(15)x_A = \frac{6375}{\tan(15^\circ)}

At point BB: tan(28)=6375xB\tan(28^\circ) = \frac{6375}{x_B} So, xB=6375tan(28)x_B = \frac{6375}{\tan(28^\circ)}

Step 3: Finding the distance the plane traveled

The plane traveled the horizontal distance between points AA and BB. So, we need to calculate xAxBx_A - x_B, the difference between the two horizontal distances.

Step 4: Calculating the values

First, we need to compute the tangent values and plug them into the equations:

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

Now, calculate the distances: xA=63750.267923883.5feetx_A = \frac{6375}{0.2679} \approx 23883.5 \, \text{feet} xB=63750.531711968.4feetx_B = \frac{6375}{0.5317} \approx 11968.4 \, \text{feet}

Step 5: Finding the traveled distance

The distance the plane traveled from point AA to point BB is: Distance=xAxB=23883.511968.4=11915.1feet\text{Distance} = x_A - x_B = 23883.5 - 11968.4 = 11915.1 \, \text{feet}

Final Answer:

The plane traveled approximately 11,915.1 feet from point AA to point BB.

Would you like any further explanation or clarification on this? Here are some related questions you might find interesting:

  1. How would the distance change if the altitude of the plane were different?
  2. How does the angle of elevation relate to the speed of the airplane?
  3. What would happen if the plane's path was not straight?
  4. How could we solve this problem using vectors instead of trigonometry?
  5. How does the tangent function change with different angles of elevation?

Tip: When working with problems involving angles and distances, always make sure your calculator is in the correct mode (degrees or radians) based on the problem.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right-Angled Triangles
Tangent Function
Angles of Elevation

Formulas

tan(θ) = height / horizontal distance

Theorems

Tangent function theorem

Suitable Grade Level

Grades 9-11