Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Ratios
A ratio is a comparison between two numbers of the same kind. It represents how many times one number contains another. It also represents how small or large one number is compared to the other.
Trigonometric Ratios
Trigonometric ratios give values of trigonometric functions. It always deals with triangles that have one angle measuring 90 degrees. These triangles are right-angled. We take the ratio of sides of these triangles.
Question
Right triangle ABC is shown in the xy-plane above. What is the value of tan A ?
Please explain, thanks!
![### Geometry Problem: Finding the Tangent of Angle \(A\)
In the diagram provided, we have a right triangle \( \triangle ABC \) situated in the \( xy \)-plane. The vertices of the triangle are given by the coordinates:
- \( A (-5, 4) \)
- \( B (7, -3) \)
- \( C (7, 4) \)
### Problem Statement
Right triangle \( \triangle ABC \) is shown in the \( xy \)-plane above. What is the value of \( \tan A \)?
### Options:
A) \( \frac{7}{12} \)
B) \( \frac{3}{4} \)
C) \( \frac{7}{9} \)
D) \( \frac{12}{7} \)
### Explanation of the Graph
In the provided graph:
- The x-axis and y-axis intersect at point \( O \) (the origin).
- Point \( A \) is located at \((-5, 4)\).
- Point \( B \) is located at \((7, -3)\).
- Point \( C \) is located at \((7, 4)\).
The triangle \( \triangle ABC \) forms a right triangle because:
- Line segment \( AC \) is horizontal.
- Line segment \( BC \) is vertical.
### Finding the Tangent of Angle \( A \)
To calculate \( \tan A \):
- First, determine the lengths of the legs \( AB \) and \( BC \).
- The length of \( BC \) is the vertical distance between \( B \) and \( C \), which is:
\[
|4 - (-3)| = 7
\]
- The length of \( AC \) is the horizontal distance between \( A \) and \( C \), which is:
\[
|7 - (-5)| = 12
\]
- \( AB \) is the hypotenuse and is not needed to calculate \( \tan A \).
The tangent of angle \( A \) is the ratio of the length of the side opposite angle \( A \) (which is \( BC \)) to the length of the side adjacent to angle \( A \) (which is \( AC \)):
\[
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5193449a-ba97-4664-aef4-702fc1ea1135%2Ff37c0d5a-37f6-4744-af37-cbbc7742c2b5%2Fznreh9r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Geometry Problem: Finding the Tangent of Angle \(A\)
In the diagram provided, we have a right triangle \( \triangle ABC \) situated in the \( xy \)-plane. The vertices of the triangle are given by the coordinates:
- \( A (-5, 4) \)
- \( B (7, -3) \)
- \( C (7, 4) \)
### Problem Statement
Right triangle \( \triangle ABC \) is shown in the \( xy \)-plane above. What is the value of \( \tan A \)?
### Options:
A) \( \frac{7}{12} \)
B) \( \frac{3}{4} \)
C) \( \frac{7}{9} \)
D) \( \frac{12}{7} \)
### Explanation of the Graph
In the provided graph:
- The x-axis and y-axis intersect at point \( O \) (the origin).
- Point \( A \) is located at \((-5, 4)\).
- Point \( B \) is located at \((7, -3)\).
- Point \( C \) is located at \((7, 4)\).
The triangle \( \triangle ABC \) forms a right triangle because:
- Line segment \( AC \) is horizontal.
- Line segment \( BC \) is vertical.
### Finding the Tangent of Angle \( A \)
To calculate \( \tan A \):
- First, determine the lengths of the legs \( AB \) and \( BC \).
- The length of \( BC \) is the vertical distance between \( B \) and \( C \), which is:
\[
|4 - (-3)| = 7
\]
- The length of \( AC \) is the horizontal distance between \( A \) and \( C \), which is:
\[
|7 - (-5)| = 12
\]
- \( AB \) is the hypotenuse and is not needed to calculate \( \tan A \).
The tangent of angle \( A \) is the ratio of the length of the side opposite angle \( A \) (which is \( BC \)) to the length of the side adjacent to angle \( A \) (which is \( AC \)):
\[
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{
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