Question

If sin θ $\frac{a}{b}$, then cos θ = ?
(a) $\frac{b}{\sqrt{b^2-a^2}}$
(b) $\frac{\sqrt{b^2-a^2}}{b}$
(c) $\frac{a}{\sqrt{b^2-a^2}}$
(d) $\frac{b}{a}$

Answer 1

(b) $\frac{\sqrt{b^2-a^2}}{b}$
Let us first draw a right $∆$ABC right angled at B and $\angle A=\theta$.
Given: sin θ = $\frac{a}{b}$, but sin θ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{a}{b}$
Thus, BC = ak and AC = bk


Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB² = (bk)² $-$ (ak)²
⇒ AB² = $\left(b^2-a^2\right)$k²
⇒ AB = ($\sqrt{b^2-a^2}$)k
∴ cos θ = $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{\sqrt{b^2-a^2}k}{bk}$ = $\frac{\sqrt{b^2-a^2}}{b}$