If \(\sin x^\circ = \frac{a}{b}\), what is \(\sin (90 - x)^\circ\)?
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
\(\sqrt{b^2 - a^3}\)
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
\(\sin x = \frac{a}{b}\)
\(\sin^{2} x + \cos^{2} x = 1\)
\(\sin^{2} x = \frac{a^{2}}{b^{2}}\)
\(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\)
\(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\)
\(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\)
= \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\)
= \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)
Source: Myschool Past Questions Archive (Record #32564)