Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
tan A = ______
I also need to know how you got the answer.
![**Using Figure 1 to Evaluate Trigonometric Functions**
In this example, we will evaluate a trigonometric function using a right triangle as depicted in Figure 1.
**Figure 1: Right Triangle Description**
Figure 1 is a right triangle, labeled as triangle \( \triangle ABC \).
- The right angle is located at point \( C \).
- Side \( BC \) is the adjacent side to angle \( A \) and measures 16 units.
- Side \( AC \) is the opposite side to angle \( A \) and measures 4 units.
To find the length of the hypotenuse \( AB \), we can use the Pythagorean theorem:
\[ AB = \sqrt{AC^2 + BC^2} = \sqrt{4^2 + 16^2} = \sqrt{16 + 256} = \sqrt{272} = 4\sqrt{17}. \]
**Evaluating Trigonometric Functions**
Assume we want to find \(\sin A\), \(\cos A\), and \(\tan A\):
- \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}}\).
- \(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}}\).
- \(\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}\).
**Note:** Remember to simplify the trigonometric values where applicable. Enter the exact answer as required.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0fb6fbe8-c93b-4e2e-8100-eedf21471e21%2Fa1a0c489-bf0b-4145-9fc8-54c0218f4b11%2F8s1ipf8_processed.png&w=3840&q=75)
Transcribed Image Text:**Using Figure 1 to Evaluate Trigonometric Functions**
In this example, we will evaluate a trigonometric function using a right triangle as depicted in Figure 1.
**Figure 1: Right Triangle Description**
Figure 1 is a right triangle, labeled as triangle \( \triangle ABC \).
- The right angle is located at point \( C \).
- Side \( BC \) is the adjacent side to angle \( A \) and measures 16 units.
- Side \( AC \) is the opposite side to angle \( A \) and measures 4 units.
To find the length of the hypotenuse \( AB \), we can use the Pythagorean theorem:
\[ AB = \sqrt{AC^2 + BC^2} = \sqrt{4^2 + 16^2} = \sqrt{16 + 256} = \sqrt{272} = 4\sqrt{17}. \]
**Evaluating Trigonometric Functions**
Assume we want to find \(\sin A\), \(\cos A\), and \(\tan A\):
- \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}}\).
- \(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}}\).
- \(\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}\).
**Note:** Remember to simplify the trigonometric values where applicable. Enter the exact answer as required.
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