88. Angle of Elevation A 28-foot building casts a 40-foot shadow on level ground. Find the angle of elevation of the sun to the nearest tenth of a degree. See the figure. 0 40 ft 28 ft
88. Angle of Elevation A 28-foot building casts a 40-foot shadow on level ground. Find the angle of elevation of the sun to the nearest tenth of a degree. See the figure. 0 40 ft 28 ft
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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![**Angle of Elevation**
A 28-foot building casts a 40-foot shadow on level ground. Find the angle of elevation \( \theta \) of the sun to the nearest tenth of a degree. See the figure.
**Explanation of Diagram:**
The diagram illustrates a right triangle formed by a building, its shadow, and the line of sight to the sun. The vertical side of the triangle represents the height of the building (28 ft), the horizontal side represents the length of the shadow (40 ft), and the hypotenuse is the line from the top of the building to the edge of the shadow, representing the angle of elevation \( \theta \).
To solve for \( \theta \), use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{28}{40}
\]
Solving for \( \theta \):
\[
\theta = \tan^{-1}\left(\frac{28}{40}\right)
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Transcribed Image Text:**Angle of Elevation**
A 28-foot building casts a 40-foot shadow on level ground. Find the angle of elevation \( \theta \) of the sun to the nearest tenth of a degree. See the figure.
**Explanation of Diagram:**
The diagram illustrates a right triangle formed by a building, its shadow, and the line of sight to the sun. The vertical side of the triangle represents the height of the building (28 ft), the horizontal side represents the length of the shadow (40 ft), and the hypotenuse is the line from the top of the building to the edge of the shadow, representing the angle of elevation \( \theta \).
To solve for \( \theta \), use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{28}{40}
\]
Solving for \( \theta \):
\[
\theta = \tan^{-1}\left(\frac{28}{40}\right)
\]
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