If \(\sin x° = \frac{a}{b}\), what is \(\sin (90 - x)°\)?
 

  • A \(\frac{\sqrt{b^2 - a^2}}{b}\)
  • B 1\(\frac{-a}{b}\)
  • C \(\frac{b^2 - a^2}{b}\)
  • D \(\frac{a^2 - b^2}{b}\)
  • E \(\sqrt{b^2 - a^3}\)
jamb 1981mathematics

The correct answer is A. \(\frac{\sqrt{b^2 - a^2}}{b}\)

If \(\sin x° = \frac{a}{b}\), then \(\sin (90 - x)° = \cos x°\). 

Using the Pythagorean identity, we know that  \(\sin^2 x° + \cos^2 x° = 1\).

Substituting the value of \(\sin x°\) into this equation,

 we get \((\frac{a}{b})^2 + \cos^2 x° = 1\). 

Solving for \(\cos x°\), we get:

\(\cos x° = \sqrt{1 - (\frac{a}{b})^2} = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}\). 

So, \(\sin (90 - x)° = \cos x° = \frac{\sqrt{b^2 - a^2}}{b}\)

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