Problem 43 - Algebra 2 and Trigonometry

Question

Jennifer lives near the airport. An airplane approaching the airport flies at a constant altitude of 1 mile toward a point, \(P,\) above Jennifer's house. Let \(\theta\) be the measure of the angle of elevation of the plane and \(d\) be the horizontal distance from \(P\) to the airplane. a. Express \(\theta\) in terms of \(d\) . b. Find \(\theta\) when \(d=1\) mile and when \(d=0.5\) mile.

Step-by-Step Solution

Answer

a) \(\theta = \tan^{-1}\left(\frac{1}{d}\right)\). b) \(\theta = 45^\circ\) for \(d=1\) mile, \(\theta \approx 63.43^\circ\) for \(d=0.5\) mile.

Step 1: Understanding the Right Triangle

Visualize the scenario as a right triangle where the airplane is at a constant height of 1 mile, the horizontal distance from the point above Jennifer's house to the airplane is \(d\), and \(\theta\) is the angle of elevation.

Step 2: Relating Angle of Elevation to Distance

The tangent of the angle of elevation \(\theta\) is the ratio of the opposite side (constant altitude = 1 mile) to the adjacent side (horizontal distance \(d\)). Therefore, \( \tan(\theta) = \frac{1}{d} \).

Step 3: Solving for Angle \(\theta\)

Express \(\theta\) in terms of \(d\) using the inverse tangent function: \( \theta = \tan^{-1}\left(\frac{1}{d}\right) \).

Step 4: Calculating \(\theta\) When \(d = 1\) Mile

Substitute \(d = 1\) into the expression for \(\theta\): \( \theta = \tan^{-1}\left(\frac{1}{1}\right) = \tan^{-1}(1) \). Since \(\tan^{-1}(1) = \frac{\pi}{4}\) radians or 45 degrees, \(\theta = 45^\circ\).

Step 5: Calculating \(\theta\) When \(d = 0.5\) Mile

Substitute \(d = 0.5\) into the expression for \(\theta\): \( \theta = \tan^{-1}\left(\frac{1}{0.5}\right) = \tan^{-1}(2) \). Use a calculator to find \(\theta \approx 63.43^\circ\).

Key Concepts

Understanding the Tangent Function

The tangent function is one of the three primary trigonometric functions. It's crucial when dealing with right triangles and is defined as the ratio of the length of the opposite side to the length of the adjacent side for a given angle. In our airplane scenario, tangent helps relate the angle of elevation, \(\theta\), to the physical dimensions of the triangle. Specifically, we have \( \tan(\theta) = \frac{1}{d} \). This beautifully demonstrates how the tangent function connects the angle \(\theta\) with the constant altitude and the horizontal distance. Let's say you have a right triangle where the height remains constant at 1 mile and the variable distance \(d\) serves as the base. The tangent function is our bridge to understanding how steeply we need to look up to see the plane.

Exploring the Right Triangle

At the heart of this problem idea is the right triangle. Right triangles are special because one angle is always 90 degrees. In our setup, the airplane, Jennifer’s house, and the point directly above her constitute a right triangle.
Your altitudinal values come from where? Let’s break it down:

• The height of the triangle is a given: 1 mile, which represents the constant altitude of the plane.
• The base is \(d\), which varies depending on how far the airplane is horizontally from point \(P\).
• \(\theta\) is the angle of elevation from point \(P\) to the airplane.

This right-angled triangle setup makes it easier to connect the sides with the functions and compute the angle of elevation. The right triangle's defined structure allows the use of trigonometric functions like tangent. This makes calculations straightforward when visualizing angles and distances.

Inverse Trigonometric Functions and Their Application

Inverse trigonometric functions allow us to find an angle when the value of a trigonometric function is given. In our case, the inverse of the tangent function, \( \tan^{-1} \), helps us find the angle \(\theta\) when the ratio \( \frac{1}{d} \) is known. This makes it immensely useful in real-life applications, like our airplane scenario, where angles are not directly measurable.
How do they work here? By setting \( \theta = \tan^{-1}\left(\frac{1}{d}\right) \), we can calculate the angle based on different values of \(d\):

• When \(d = 1\) mile, \(\theta = \tan^{-1}(1) = 45^\circ\).
• When \(d = 0.5\) mile, it's approximately \(\theta = 63.43^\circ\) by using a calculator for \( \tan^{-1}(2) \).

This process showcases the power of inverse functions: turning known ratios into useful angle measurements.

Navigating Constant Altitude

In problems where movement occurs but height remains the same, constant altitude is a key concept. In our scenario, the airplane flies at a constant altitude of 1 mile.
What does constant altitude imply?

• The height of 1 mile stays the same, unaffected by how far horizontally the plane is from \(P\).
• This sets a stable frame of reference for measuring the angle \(\theta\).
• It ensures a consistent triangle height no matter how \(d\) changes.

Having a consistent altitude simplifies calculations: only the base \(d\) and angle \(\theta\) vary. It stabilizes one side of your right triangle, leaving the other two to fluctuate, allowing you to effectively use trigonometric functions during your computations. Understanding constant altitude is essential because it serves as a fixed point in many real-world applications involving flight paths or line-of-sight problems.