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

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}

Correct answer is A

\sin x = \frac{a}{b}

\sin^{2} x + \cos^{2} x = 1

\sin^{2} x = \frac{a^{2}}{b^{2}}

\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}

\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}

\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x

= (1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})

= \frac{\sqrt{b^{2} - a^{2}}}{b}