Two guy wires are attached to utility poles that are 40 ft apart and joined at the same point on the ground, as shown in the following figure: 30 ft 20 ft We will call the lengths of the two guy wires h1 and h2, respectively. Further, notice that the distance along the ground is 40 ft and is broken up into two segments: x and y (x is already labeled on the picture; y is the other segment along the ground). For this problem, we will assume that x = 30 ft. Blank 1: Find the value of y. Blank 2: Find the value, rounded to the nearest tenth of a foot, of h1 + h2.

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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### Geometry Problem: Guy Wires and Utility Poles

#### Problem Statement
Two guy wires are attached to utility poles that are 40 ft apart and joined at the same point on the ground, as shown in the following figure:

![Figure depicting two utility poles with guy wires](https://example.com/diagram.jpg)

The diagram shows two utility poles of different heights, 20 ft and 30 ft, each with a guy wire attached and meeting at a point on the ground. The distance between the two poles along the ground is 40 ft.

#### Definitions and Variables
The lengths of the two guy wires are denoted as \( h_1 \) and \( h_2 \), respectively. The ground distance of 40 ft is divided into two segments: \( x \) and \( y \).

- \( x \): Distance along the ground between one of the utility poles and the point where the guy wires meet (given as 30 ft).
- \( y \): The remaining distance along the ground between the other utility pole and the point where the guy wires meet. 
- \( h_1, h_2 \): Lengths of the guy wires.

#### Given
- Distance between poles: 40 ft
- \( x \) is given as 30 ft

#### Problems to Solve
**Blank 1:** Find the value of \( y \).

**Blank 2:** Find the value, rounded to the nearest tenth of a foot, of \( h_1 + h_2 \).

#### Solution Approach
Using the Pythagorean Theorem, the lengths of the guy wires \( h_1 \) and \( h_2 \) can be determined:
 
\[ h_1 = \sqrt{(20)^2 + (30)^2} \]
\[ h_2 = \sqrt{(30)^2 + (10)^2} \]

where:
- \( h_1 \) is the hypotenuse formed by the 20 ft pole and 30 ft ground segment.
- \( h_2 \) is the hypotenuse formed by the 30 ft pole and the remaining 10 ft ground segment \( y \), as the total horizontal distance must equal the given 40 ft.

By solving these, \( h_1 \) and \( h_2 \) can be added together to find \( h_1 + h_2 \).

This problem offers a practical application of the
Transcribed Image Text:### Geometry Problem: Guy Wires and Utility Poles #### Problem Statement Two guy wires are attached to utility poles that are 40 ft apart and joined at the same point on the ground, as shown in the following figure: ![Figure depicting two utility poles with guy wires](https://example.com/diagram.jpg) The diagram shows two utility poles of different heights, 20 ft and 30 ft, each with a guy wire attached and meeting at a point on the ground. The distance between the two poles along the ground is 40 ft. #### Definitions and Variables The lengths of the two guy wires are denoted as \( h_1 \) and \( h_2 \), respectively. The ground distance of 40 ft is divided into two segments: \( x \) and \( y \). - \( x \): Distance along the ground between one of the utility poles and the point where the guy wires meet (given as 30 ft). - \( y \): The remaining distance along the ground between the other utility pole and the point where the guy wires meet. - \( h_1, h_2 \): Lengths of the guy wires. #### Given - Distance between poles: 40 ft - \( x \) is given as 30 ft #### Problems to Solve **Blank 1:** Find the value of \( y \). **Blank 2:** Find the value, rounded to the nearest tenth of a foot, of \( h_1 + h_2 \). #### Solution Approach Using the Pythagorean Theorem, the lengths of the guy wires \( h_1 \) and \( h_2 \) can be determined: \[ h_1 = \sqrt{(20)^2 + (30)^2} \] \[ h_2 = \sqrt{(30)^2 + (10)^2} \] where: - \( h_1 \) is the hypotenuse formed by the 20 ft pole and 30 ft ground segment. - \( h_2 \) is the hypotenuse formed by the 30 ft pole and the remaining 10 ft ground segment \( y \), as the total horizontal distance must equal the given 40 ft. By solving these, \( h_1 \) and \( h_2 \) can be added together to find \( h_1 + h_2 \). This problem offers a practical application of the
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