Problem 61 - Algebra and Trigonometry

Question

Height of a Rocket A rocket fired straight up is tracked by an observer on the ground a mile away. (a) Show that when the angle of elevation is \(\theta,\) the height of the rocket in feet is \(h=5280 \tan \theta\) (b) Complete the table to find the height of the rocket at the given angles of elevation. $$\begin{array}{|c|c|c|c|c|c|}\hline \theta & {20^{\circ}} & {60^{\circ}} & {80^{\circ}} & {85^{\circ}} \\ \hline h & {} & {} & {} \\\ \hline\end{array}$$

Step-by-Step Solution

Answer

The height formula is \(h=5280 \tan \theta\) and calculated heights are 1920.32 ft, 9137.76 ft, 29921.58 ft, and 60358.13 ft.

Step 1: Understand the Problem

We are given an observer a mile away tracking a rocket fired straight up and need to find a relationship between the angle of elevation \(\theta\) and the height of the rocket \(h\) in feet.

Step 2: Establish the Relationship

Since the observer is 1 mile (5280 feet) away from the launch point, we form a right triangle where the adjacent side is 5280 feet and the opposite side is the height of the rocket \(h\). Using the tangent function, \(\tan \theta = \frac{h}{5280}\), we can rearrange this to \(h = 5280 \tan \theta\). This equation gives the height of the rocket in terms of \(\theta\).

Step 3: Evaluate for \(20^\circ\)

Substitute \(\theta = 20^\circ\) into the formula: \(h = 5280 \tan 20^\circ\). Calculate \(\tan 20^\circ\) using a calculator to find \(h \approx 5280 \times 0.3640 \approx 1920.32\) feet.

Step 4: Evaluate for \(60^\circ\)

Substitute \(\theta = 60^\circ\) into the equation: \(h = 5280 \tan 60^\circ\). Knowing \(\tan 60^\circ = \sqrt{3} \approx 1.732\), calculate \(h \approx 5280 \times 1.732 \approx 9137.76\) feet.

Step 5: Evaluate for \(80^\circ\)

Substitute \(\theta = 80^\circ\): \(h = 5280 \tan 80^\circ\). Daycalculator gives \(\tan 80^\circ \approx 5.6713\), hence \(h \approx 5280 \times 5.6713 \approx 29921.58\) feet.

Step 6: Evaluate for \(85^\circ\)

Calculate \(h\) with \(\theta = 85^\circ\): \(h = 5280 \tan 85^\circ\). Using \(\tan 85^\circ \approx 11.4301\), we get \(h \approx 5280 \times 11.4301 \approx 60358.13\) feet.

Step 7: Complete the Table

Fill in the heights calculated for each angle into the table:\[\begin{array}{|c|c|c|c|c|c|}\hline\theta & 20^{\circ} & 60^{\circ} & 80^{\circ} & 85^{\circ} \\hlineh & 1920.32 & 9137.76 & 29921.58 & 60358.13 \\hline\end{array}\]

Key Concepts

Understanding the Angle of Elevation

The angle of elevation is a key concept in trigonometry and is defined as the angle between the line of sight and the horizontal plane when an observer is looking at an object above the horizontal level. In this exercise, the observer is stationed at a given distance from the base of an object, such as a rocket. Since the rocket is launched vertically, the observer looks upward, forming an angle with their line of sight relative to the ground. This angle helps determine the height of the object using trigonometric relationships.

It’s important to remember:

• The angle of elevation is always measured from the horizontal line up to the line of sight.
• This angle plays a crucial role in calculating heights and distances in problems involving heights, depressions, and several fields such as navigation and architecture.

The angle of elevation helps convert the distance components of the object into a precise height measurement with the help of trigonometric functions.

The Tangent Function in Trigonometry

The tangent function is one of the fundamental trigonometric functions and is essential for dealing with relationships in right triangles. It relates the opposite side to the adjacent side in a right triangle. The formula is:

\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \]
In the context of this problem, the tangent function is used to link the height of the rocket to its distance from the observer. Here, the height (\( h \)) is the opposite side, and the distance from the observer (\( 5280 \, \text{feet} \)) is the adjacent side. Therefore, using the tangent function, we can express the height of the rocket as:

\[ h = 5280 \tan \theta \]
This relationship becomes extremely useful when you plug in various values of \( \theta \) (the angle of elevation) to compute the height of objects such as rockets or tall buildings. By exploiting this trigonometric function, you can solve many practical problems involving heights and distances.

Exploring the Right Triangle

Right triangles hold a special place in trigonometry. They have one angle measuring 90 degrees, and the other two angles sum up to 90 degrees as well. This is essential when determining angles like the angle of elevation in real-world scenarios.

In our scenario, the right triangle is formed with the following components:

• The adjacent side, which is the horizontal distance from the observer to the launch point (5280 feet).
• The opposite side, which is the vertical height of the rocket from the ground (\( h \)).
• The hypotenuse, which is not directly used in this specific case but is the longest side opposite the right angle.

By forming a right triangle, applying trigonometric functions like tangent becomes possible. This geometric structure allows us to apply the equation \( h = 5280 \tan \theta \). Right triangles are a cornerstone in solving many geometric and real-world problems efficiently and consist of consistent rules that reliably solve wide-ranging issues.

Calculating the Height of the Rocket

To calculate the height of the rocket at various angles of elevation, the equation \( h = 5280 \tan \theta \) is used. This equation is derived from the tangent function specific to the problem setup and measurements noted.

Depending on the angle of elevation, the tangent value changes:

• For smaller angles, the tangent is relatively small, leading to a lower calculated height.
• As the angle approaches 90 degrees, the tangent and thus the height becomes very large.

Let's break it down for some angles:

- At 20°, the calculation is \( h = 5280 \tan 20° \approx 1920.32 \, \text{feet} \)
- For 60°, it becomes \( h = 5280 \tan 60° \approx 9137.76 \, \text{feet} \)
- At 80°, we see \( h = 5280 \tan 80° \approx 29921.58 \, \text{feet} \)
- Finally, for 85°, \( h = 5280 \tan 85° \approx 60358.13 \, \text{feet} \)

This height calculation provides a clear picture of how the height changes with varying angles and demonstrates the power of trigonometry in practical applications.