We are given that $sinA = \frac{m}{n}$ and the angle A lies in the first quadrant, i.e., $A \in (0, \frac{\pi}{2})$. We need to find the value of $cosA$.

Trigonometric Identity Application

We use the fundamental trigonometric identity:

$sin^2A + cos^2A = 1$

Substituting Given Value

Substitute the given value of $sinA$ into the identity:

$(\frac{m}{n})^2 + cos^2A = 1$

$\frac{m^2}{n^2} + cos^2A = 1$

Solving for $cos^2A$

Rearrange the equation to solve for $cos^2A$:

$cos^2A = 1 - \frac{m^2}{n^2}$

$cos^2A = \frac{n^2 - m^2}{n^2}$

Finding $cosA$

Take the square root of both sides:

$cosA = \pm \sqrt{\frac{n^2 - m^2}{n^2}}$

$cosA = \pm \frac{\sqrt{n^2 - m^2}}{n}$

Determining the Sign of $cosA$

Since A is in the first quadrant ($A \in (0, \frac{\pi}{2})$), the value of $cosA$ is positive.

Therefore, we take the positive square root:

$cosA = \frac{\sqrt{n^2 - m^2}}{n}$