Given that $\alpha$ is an acute angle and $\tan(\alpha) = \frac{4}{3}$, we want to find $\cos(\alpha)$.

The trigonometric identity that relates $\tan(\alpha)$ to $\sin(\alpha)$ and $\cos(\alpha)$ is: $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}.$ Since $\tan(\alpha) = \frac{4}{3}$, this implies: $\frac{\sin(\alpha)}{\cos(\alpha)} = \frac{4}{3}.$ Let $\sin(\alpha) = 4k$ and $\cos(\alpha) = 3k$. To find $k$, we use the Pythagorean identity: $\sin^2(\alpha) + \cos^2(\alpha) = 1.$ Substitute $\sin(\alpha)$ and $\cos(\alpha)$: $(4k)^2 + (3k)^2 = 1,$ $16k^2 + 9k^2 = 1,$ $25k^2 = 1.$ Solve for $k^2$: $k^2 = \frac{1}{25}.$ Taking the positive square root (since $\alpha$ is acute and both $\sin$ and $\cos$ are positive in the first quadrant): $k = \frac{1}{5}.$

Now, substitute $k$ back into the expressions for $\sin(\alpha)$ and $\cos(\alpha)$: $\sin(\alpha) = 4k = 4 \times \frac{1}{5} = \frac{4}{5},$ $\cos(\alpha) = 3k = 3 \times \frac{1}{5} = \frac{3}{5}.$

Thus, $\cos(\alpha) = \frac{3}{5}$.

Do you want more details or have any questions?

Here are 5 related questions:

1. What is $\sin(\alpha)$ given $\tan(\alpha) = \frac{4}{3}$?
2. How can you find $\sec(\alpha)$ using $\tan(\alpha)$?
3. What is the value of $\cot(\alpha)$ for this angle?
4. How does knowing the quadrant of an angle affect the signs of trigonometric functions?
5. How do you derive the Pythagorean identity?

Tip: Always ensure to identify the quadrant of an angle, as it determines the signs of the trigonometric functions.