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Chapter 3: Problem 13

Looking over a wall: Twenty horizontal feet north of a 50 -foot building is a 35 -foot wall (see Figure 3.37). A man 6 feet tall wishes to view the top of the building from the north side of the wall. How far north of the wall must he stand in order to view the top of the building?

Short Answer

Expert verified
The man must stand approximately 38.67 feet north of the wall.

Step by step solution

01

Define the Geometry of the Problem

Visualize the scenario as a right triangle problem. The man's eyes are at a height of 6 feet, the wall is 35 feet tall and is 20 horizontal feet from the building. The building is 50 feet tall.
02

Calculate the Optical Line

The man wants to view the top of the building, which forms an imaginary line from his eyes to the top of the building. To just clear the wall, this line must be tangent to the top of the wall.
03

Set Up the Problem Using Similar Triangles

There are two triangles to consider: one small triangle from the top of the wall to the top of the building, and a larger triangle from the man's eyes to the top of the building. Both triangles will share an angle at the top of the building, making them similar triangles.
04

Write Equation Using Similar Triangles

The height from the wall to the top of the building is 15 feet (i.e., 50 - 35). Let the distance the man stands north of the wall be denoted as x. The triangles give the ratio: \( \frac{15}{20} = \frac{44}{x+20} \).
05

Solve the Ratio Equation for x

Simplify and solve the equation: \[ \frac{15}{20} = \frac{44}{x+20} \] Multiply across the relationship to eliminate fractions: \( 15(x + 20) = 44 \times 20 \). \( 15x + 300 = 880\). \( 15x = 580 \). \( x = \frac{580}{15} \). \( x \approx 38.67 \) feet.
06

Review and Verify Solution

Verify that when the man stands approximately 38.67 feet north of the wall, the line of sight from his eyes to the top of the building just clears the wall. This confirms the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangles
In the context of this exercise, a right triangle is key in helping visualize the relationship between different heights and distances. A right triangle has one 90-degree angle and the other two angles make up 90 degrees together. Right triangles are often used in problems involving height and distance because they are easy to work with mathematically.

In our problem, the right triangle is formed by
  • The horizontal distance between the wall and the building (20 feet).
  • The vertical height difference between the top of the wall and the top of the building (15 feet, which is calculated as 50 feet for the building minus 35 feet for the wall).


This triangle helps define the optical line along which the man would look to see the top of the building over the wall. Understanding this concept helps you solve problems in which you need to find missing sides or angles, given all the other elements of the triangle.
Similar Triangles
Understanding similar triangles is crucial in this problem because it involves two triangles that share common angles. Similar triangles are triangles that have the same shape but may have different sizes. This means that their corresponding angles are equal and their sides are proportional.

In our scenario, we have:
  • A smaller triangle that goes from the top of the wall to the top of the building.
  • A larger triangle formed by the line from the man's eyes to the top of the building.

The point where these two triangles share an angle is at the top of the building. Because of this shared angle, we know the triangles are similar. We can then set up a proportion between the sides of these triangles to solve for missing values like the distance the man must stand north of the wall to just see over the wall. These proportions help us develop equations that we can solve for unknowns.
Problem Solving
Problem solving using trigonometric concepts like right triangles and similar triangles requires a strategic approach. Here is a step-by-step strategy used in solving such problems:
  • Start by visualizing the situation and identify all given information.
  • Recognize the geometric shapes within the problem, like triangles, and determine how they relate to one another.
  • Apply geometric principles such as the properties of right triangles and the rules of similar triangles.
  • Set up equations based on these properties and solve for the unknown values.
  • Finally, verify the solution to ensure it satisfies the conditions laid out in the problem statement.

In this particular exercise, the problem-solving process involves understanding the real-world situation (looking over the wall) in terms of mathematical shapes (triangles). By using proportions from similar triangles, we can create an equation that helps identify how far the man must stand from the wall. Once solved, reviewing the solution ensures the calculation accurately addresses the problem. This systematic approach is invaluable in tackling trigonometry problems effectively.

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Most popular questions from this chapter

Expansion of steam: When water changes to steam, its volume increases rapidly. At a normal atmospheric pressure of \(14.7\) pounds per square inch, water boils at 212 degrees Fahrenheit and expands in volume by a factor of 1700 to 1 . But when water is sprayed into hotter areas, the expansion ratio is much greater. This principle can be applied to good effect in fire fighting. The steam can occupy such a large volume that oxygen is expelled from the area and the fire may be smothered. The table below shows the approximate volume, in cubic feet, of 50 gallons of water converted to steam at the given temperatures, in degrees Fahrenheit. $$ \begin{array}{|c|c|} \hline T=\text { Temperature } & V=\text { cubic } \\ \text { feet of steam } \\ \hline 212 & 10,000 \\ \hline 400 & 12,500 \\ \hline 500 & 14,100 \\ \hline 800 & 17,500 \\ \hline 1000 & 20,000 \\ \hline \end{array} $$ a. Make a linear model of volume \(V\) as a function of \(T\). b. If one fire is 100 degrees hotter than another, what is the increase in the volume of steam produced by 50 gallons of water? c. Calculate \(V(420)\) and explain in practical terms what your answer means. d. At a certain fire, 50 gallons of water expanded to 14,200 cubic feet of steam. What was the temperature of the fire?

An application of three equations in three unknowns: A bag of coins contains nickels, dimes, and quarters. There are a total of 21 coins in the bag, and the total amount of money in the bag is \(\$ 3.35\). There is one more dime than there are nickels. How many dimes, nickels, and quarters are in the bag?

Market supply and demand: The quantity of wheat, in billions of bushels, that wheat suppliers are willing to produce in a year and offer for sale is called the quantity supplied and is denoted by \(S\). The quantity supplied is determined by the price \(P\) of wheat, in dollars per bushel, and the relation is \(P=2.13 S-0.75 .\) The quantity of wheat, in billions of bushels, that wheat consumers are willing to purchase in a year is called the quantity demanded and is denoted by \(D\). The quantity demanded is also determined by the price \(P\) of wheat, and the relation is \(P=2.65-0.55 D\). At the equilibrium price, the quantity supplied and the quantity demanded are the same. Find the equilibrium price for wheat.

Sports car: An automotive engineer is testing how quickly a newly designed sports car accelerates from rest. He has collected data giving the velocity, in miles per hour, as a function of the time, in seconds, since the car was at rest. Here is a table giving a portion of his data: \begin{tabular}{|c|c|} \hline Time & Velocity \\ \hline \(2.0\) & \(27.9\) \\ \hline \(2.5\) & \(33.8\) \\ \hline \(3.0\) & \(39.7\) \\ \hline \(3.5\) & \(45.6\) \\ \hline \end{tabular} a. By calculating differences, show that the data in this table can be modeled by a linear function. b. What is the slope for the linear function modeling velocity as a function of time? Explain in practical terms the meaning of the slope. c. Use the data in the table to find the formula for velocity as a linear function of time that is valid over the time period covered in the table. d. What would your formula from part \(\mathrm{c}\) give for the velocity of the car at time 0 ? What does this say about the validity of the linear formula over the initial segment of the experiment? Explain your answer in practical terms. e. Assume that the linear formula you found in part \(\mathrm{c}\) is valid from 2 seconds through 5 seconds. The marketing department wants to know from the engineer how to complete the following statement: "This car goes from 0 to \(60 \mathrm{mph}\) in seconds." How should they fill in the blank?

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