Question

If \( \sin \theta = \frac{a}{b} \), then the value of \( \cos \theta \) is:

• A. \( \frac{b}{\sqrt{b^2 - a^2}} \)
• B. \( \frac{\sqrt{b^2 - a^2}}{b} \)
• C. \( \frac{a}{\sqrt{b^2 - a^2}} \)
• D. \( \frac{b}{a} \)

Hint:

Use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to find \( \cos \theta \) when \( \sin \theta \) is known.

Answer 1

The Correct Option is B

We are given that \( \sin \theta = \frac{a}{b} \), where \( b \) is the hypotenuse and \( a \) is the side opposite to the angle \( \theta \). To find \( \cos \theta \), we can use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \] Substitute \( \sin \theta = \frac{a}{b} \) into the identity: \[ \left( \frac{a}{b} \right)^2 + \cos^2 \theta = 1 \quad \Rightarrow \quad \frac{a^2}{b^2} + \cos^2 \theta = 1. \] Solving for \( \cos^2 \theta \): \[ \cos^2 \theta = 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2}. \] Thus, \( \cos \theta = \frac{\sqrt{b^2 - a^2}}{b} \). Therefore, the value of \( \cos \theta \) is \( \boxed{\frac{\sqrt{b^2 - a^2}}{b}} \).