Let a and b be positive numbers. Evaluate sin (tan O 60 b a² +6² a (tan-¹ (7)). √a² +6² √a² +6² a² +6² /a² +6² b
Let a and b be positive numbers. Evaluate sin (tan O 60 b a² +6² a (tan-¹ (7)). √a² +6² √a² +6² a² +6² /a² +6² b
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°.
Related questions
Question
![**Problem Statement:**
Let \( a \) and \( b \) be positive numbers. Evaluate \( \sin \left( \tan^{-1} \left( \frac{a}{b} \right) \right) \).
**Options:**
- \( \frac{b}{\sqrt{a^2 + b^2}} \)
- \( \frac{b}{a} \)
- \( \frac{a}{\sqrt{a^2 + b^2}} \) (Selected answer)
- \( \frac{\sqrt{a^2 + b^2}}{a^2 + b^2} \)
- \( \frac{\sqrt{a^2 + b^2}}{b} \)
**Explanation:**
The problem involves finding the sine of the angle whose tangent is the ratio \( \frac{a}{b} \). To solve this, consider a right triangle where \( a \) is the opposite side and \( b \) is the adjacent side. The hypotenuse of the triangle can be found using the Pythagorean theorem: \( \sqrt{a^2 + b^2} \).
Thus:
\[ \sin \left( \tan^{-1} \left( \frac{a}{b} \right) \right) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{\sqrt{a^2 + b^2}} \]
So, the correct option is:
\[ \frac{a}{\sqrt{a^2 + b^2}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c85a724-3a2d-442a-921e-64e5cff25e84%2F28a4bade-fc3b-43dd-b0d4-5af4202ec69a%2Fu8vlg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( a \) and \( b \) be positive numbers. Evaluate \( \sin \left( \tan^{-1} \left( \frac{a}{b} \right) \right) \).
**Options:**
- \( \frac{b}{\sqrt{a^2 + b^2}} \)
- \( \frac{b}{a} \)
- \( \frac{a}{\sqrt{a^2 + b^2}} \) (Selected answer)
- \( \frac{\sqrt{a^2 + b^2}}{a^2 + b^2} \)
- \( \frac{\sqrt{a^2 + b^2}}{b} \)
**Explanation:**
The problem involves finding the sine of the angle whose tangent is the ratio \( \frac{a}{b} \). To solve this, consider a right triangle where \( a \) is the opposite side and \( b \) is the adjacent side. The hypotenuse of the triangle can be found using the Pythagorean theorem: \( \sqrt{a^2 + b^2} \).
Thus:
\[ \sin \left( \tan^{-1} \left( \frac{a}{b} \right) \right) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{\sqrt{a^2 + b^2}} \]
So, the correct option is:
\[ \frac{a}{\sqrt{a^2 + b^2}} \]
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