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If \(\sin \theta = a/b\), then \(\cos \theta\) is equal to:

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

Answer: C

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\((\frac{a}{b})^2 + \cos^2 \theta = 1\)

\(\cos^2 \theta = 1 - \frac{a^2}{b^2} = \frac{b^2-a^2}{b^2}\)

\(\cos \theta = \sqrt{\frac{b^2-a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}\).

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