Given that sin θ = $\Big(\dfrac{a}{b}\Big)$, then cos θ is equal to:

1. $\Big(\dfrac{b}{a}\Big)$

2. $\Big(\dfrac{a}{\sqrt{b^2 - a^2}}\Big)$

3. $\Big(\dfrac{b}{\sqrt{b^2 - a^2}}\Big)$

4. $\Big(\dfrac{\sqrt{b^2 - a^2}}{b}\Big)$

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

By formula,

$\sin \theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}}$

Substituting values we get :

$\dfrac{a}{b} = \dfrac{AB}{AC}$

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (bk)² = (ak)² + BC²

⇒ b²k² = a²k² + BC²

⇒ BC² = b²k² - a²k²

⇒ BC = k$\sqrt{b^2 - a^2}$

By formula,

$\cos \theta = \dfrac{\text{base}}{\text{hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{b^2 - a^2}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b} .$

Hence, option 4 is the correct option.