Question

Given that $\sin\theta = \frac{a}{b}$, then $\cos\theta$ is equal to :

• A. $\frac{b}{\sqrt{b^2-a^2}}$
• B. $\frac{b}{a}$
• C. $\frac{\sqrt{b^2-a^2}}{b}$
• D. $\frac{a}{\sqrt{b^2-a^2}}$

Hint:

Using a quick mental right-triangle is often faster:
If Opposite $= a$ and Hypotenuse $= b$, then by Pythagoras, the Adjacent side is $\sqrt{b^2 - a^2}$.
Since cosine is Adjacent / Hypotenuse, we immediately get $\frac{\sqrt{b^2 - a^2}}{b}$.

Answer 1

The Correct Option is C

Step 1: Model it as a right triangle instead of using the identity directly.
Since $\sin\theta=\frac{a}{b}$, picture a right triangle where the side opposite $\theta$ is $a$ and the hypotenuse is $b$.
Step 2: Find the adjacent side with Pythagoras.
\[ \text{Adjacent} = \sqrt{b^2-a^2} \]
Step 3: Read off cosine from the triangle.
$\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{b^2-a^2}}{b}$.
\[ \boxed{\dfrac{\sqrt{b^2-a^2}}{b}} \]