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To solve for \(\sin \theta\) and \(\tan \theta\) given that \(\cos \theta = \frac{5}{13}\) and \(0 \leq \theta < \frac{\pi}{2}\), we can use the Pythagorean identity and the definition of tangent:

### Step 1: Find \(\sin \theta\)
We know the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substitute \(\cos \theta = \frac{5}{13}\):
\[
\sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1
\]
\[
\sin^2 \theta + \frac{25}{169} = 1
\]
\[
\sin^2 \theta = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}
\]
\[
\sin \theta = \sqrt{\frac{144}{169}} = \frac{12}{13}
\]
Since \(\theta\) is in the first quadrant (\(0 \leq \theta < \frac{\pi}{2}\)), \(\sin \theta\) is positive, so:
\[
\sin \theta = \frac{12}{13}
\]

### Step 2: Find \(\tan \theta\)
The tangent of an angle is given by:
\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\]
Substitute \(\sin \theta = \frac{12}{13}\) and \(\cos \theta = \frac{5}{13}\):
\[
\tan \theta = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}
\]

### Final Answers:
- \(\sin \theta = \frac{12}{13}\)
- \(\tan \theta = \frac{12}{5}\)

Would you like further details or have any questions? 

Here are some related questions for practice:
1. What is \(\cos \theta\) if \(\sin \theta = \frac{7}{25}\) and \(\theta\) is in the first quadrant?
2. If \(\tan \theta = 3\), what are the values of \(\sin \theta\) and \(\cos \theta\)?
3. Can you find \(\sin \theta\) and \(\cos \theta\) if \(\tan \theta = \frac{3}{4}\)?
4. How does the identity \(\sin^2 \theta + \cos^2 \theta = 1\) help in solving trigonometric equations?
5. What is the value of \(\tan \theta\) if \(\cos \theta = \frac{8}{17}\)?

Tip: Always check which quadrant the angle belongs to when determining signs for \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\).

Find sin(θ) and tan(θ) if cos(θ) = 5/13, assuming that 0 ≤ θ < π/2.

Learn how to find sin(θ) and tan(θ) when cos(θ) = 5/13 in the first quadrant (0 ≤ θ < π/2). This solution uses key trigonometric formulas, including the Pythagorean identity and the definition of tangent. Suitable for students in Grades 9-11.

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