Step 1: Build an actual right triangle instead of squaring the identity.
Since $\sin \theta$ is a ratio of sides in a right triangle, let's draw that triangle and label its sides directly, rather than manipulating the Pythagorean identity algebraically.
Step 2: Label the triangle using the given ratio.
In a right angled triangle with one acute angle $\theta$:
\[ \sin \theta = \frac{\text{side opposite } \theta}{\text{hypotenuse}} \]
We are given $\sin \theta = \frac{a}{b}$, so we can directly say the side opposite $\theta$ has length $a$ and the hypotenuse has length $b$.
Step 3: Find the third side using Pythagoras' theorem.
Let the side adjacent to $\theta$ (the base) have length $x$. By Pythagoras' theorem in this right triangle:
\[ (\text{hypotenuse})^2 = (\text{opposite})^2 + (\text{adjacent})^2 \]
\[ b^2 = a^2 + x^2 \]
Solve for $x$:
\[ x^2 = b^2 - a^2 \implies x = \sqrt{b^2 - a^2} \]
Step 4: Read off $\cos \theta$ from the triangle.
By definition, $\cos \theta$ is the adjacent side over the hypotenuse:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{b^2 - a^2}}{b} \]
Step 5: Final answer.
$\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}$, which is option (C).
\[ \boxed{\cos \theta = \dfrac{\sqrt{b^2 - a^2}}{b}} \]