Suppose alpha is an acute angle and tan(alpha) = 4/3. What is cos(alpha)?

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.

Learn how to calculate the cosine of an acute angle when given that tan(α) = 4/3. This step-by-step solution applies trigonometric identities, including the Pythagorean identity, to find cos(α). Suitable for students in Grades 9-12.

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