A radio station tower was built in two sections. From a point 82 feet from the base of the tower, the angle of elevation to the top of the lower section is 35°, and the angle of elevation to the top of the upper section is 48°. Round your answers to the nearest foot. How tall is the tower? feet. How tall is the lower section of the tower? How tall is the upper section of the tower? feet. feet.
A radio station tower was built in two sections. From a point 82 feet from the base of the tower, the angle of elevation to the top of the lower section is 35°, and the angle of elevation to the top of the upper section is 48°. Round your answers to the nearest foot. How tall is the tower? feet. How tall is the lower section of the tower? How tall is the upper section of the tower? feet. feet.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![A radio station tower was built in two sections. From a point 82 feet from the base of the tower, the angle of elevation to the top of the lower section is \(35^\circ\), and the angle of elevation to the top of the upper section is \(48^\circ\).
**Round your answers to the nearest foot.**
1. **How tall is the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
2. **How tall is the lower section of the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
3. **How tall is the upper section of the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
To solve these problems, use trigonometric functions, especially the tangent function given the angle of elevation and the distance from the point to the base of the tower.
Detailed diagram explanation:
The problem does not provide a specific diagram, but a typical illustration for this problem would consist of:
- A horizontal line representing the ground.
- A vertical line from the base of the tower to represent the height of the lower section and the upper section.
- A point to the right of the base, 82 feet away, forming the adjacent side of right-angled triangles with the lower and upper sections of the tower.
For each section, draw:
- A right-angled triangle where the base (adjacent side) is 82 feet.
- An angle of elevation for that section (\(35^\circ\) for the lower section and \(48^\circ\) for the upper section).
- The opposite side of each triangle represents the height of each section.
By these constructions, you can use the formula for tangent (\(\tan(\theta) = \text{opposite}/\text{adjacent}\)) to find the heights.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6aa8815b-de7d-43a5-aa48-91a8705759b5%2Fb985e2d4-9b15-4916-9130-1c3a6ae34e9a%2Falhnohe_processed.png&w=3840&q=75)
Transcribed Image Text:A radio station tower was built in two sections. From a point 82 feet from the base of the tower, the angle of elevation to the top of the lower section is \(35^\circ\), and the angle of elevation to the top of the upper section is \(48^\circ\).
**Round your answers to the nearest foot.**
1. **How tall is the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
2. **How tall is the lower section of the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
3. **How tall is the upper section of the tower?**
\[ \_\_\_\_\_\_ \text{ feet.} \]
To solve these problems, use trigonometric functions, especially the tangent function given the angle of elevation and the distance from the point to the base of the tower.
Detailed diagram explanation:
The problem does not provide a specific diagram, but a typical illustration for this problem would consist of:
- A horizontal line representing the ground.
- A vertical line from the base of the tower to represent the height of the lower section and the upper section.
- A point to the right of the base, 82 feet away, forming the adjacent side of right-angled triangles with the lower and upper sections of the tower.
For each section, draw:
- A right-angled triangle where the base (adjacent side) is 82 feet.
- An angle of elevation for that section (\(35^\circ\) for the lower section and \(48^\circ\) for the upper section).
- The opposite side of each triangle represents the height of each section.
By these constructions, you can use the formula for tangent (\(\tan(\theta) = \text{opposite}/\text{adjacent}\)) to find the heights.
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