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

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

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

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

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

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

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

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

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