Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
Bartleby Related Questions Icon

Related questions

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.
expand button
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.
Expert Solution
Check Mark
Step 1

Calculus homework question answer, step 1, image 1

Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Calculus: Early Transcendentals
Calculus
ISBN:9781285741550
Author:James Stewart
Publisher:Cengage Learning
Text book image
Thomas' Calculus (14th Edition)
Calculus
ISBN:9780134438986
Author:Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:PEARSON
Text book image
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:9780134763644
Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:PEARSON
Text book image
Calculus: Early Transcendentals
Calculus
ISBN:9781319050740
Author:Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:W. H. Freeman
Text book image
Precalculus
Calculus
ISBN:9780135189405
Author:Michael Sullivan
Publisher:PEARSON
Text book image
Calculus: Early Transcendental Functions
Calculus
ISBN:9781337552516
Author:Ron Larson, Bruce H. Edwards
Publisher:Cengage Learning