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We are given that \(\sin x = \frac{5}{13}\) and that \(90^\circ \leq x \leq 180^\circ\). We are asked to find \(\cos x\).

### Step-by-step Solution:
1. Since we know that \(90^\circ \leq x \leq 180^\circ\), this means that \(x\) is in the second quadrant. In the second quadrant, sine is positive, and cosine is negative.

2. We can use the Pythagorean identity to find \(\cos x\):
   \[
   \sin^2 x + \cos^2 x = 1
   \]
   Substituting \(\sin x = \frac{5}{13}\) into this equation:
   \[
   \left(\frac{5}{13}\right)^2 + \cos^2 x = 1
   \]
   \[
   \frac{25}{169} + \cos^2 x = 1
   \]
   Solving for \(\cos^2 x\):
   \[
   \cos^2 x = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}
   \]
   Taking the square root of both sides:
   \[
   \cos x = \pm \frac{12}{13}
   \]
   
3. Since \(x\) is in the second quadrant, \(\cos x\) must be negative, so:
   \[
   \cos x = -\frac{12}{13}
   \]

### Answer:
The correct value of \(\cos x\) is \(-\frac{12}{13}\), which corresponds to option A.

Would you like further clarification or more details on this solution? Here are some related questions to expand on this topic:

1. What is the value of \(\tan x\) in this scenario?
2. Can you explain the behavior of sine, cosine, and tangent in each quadrant?
3. How do we find the angle \(x\) if given \(\sin x\)?
4. What is the general Pythagorean identity for trigonometric functions?
5. How does cosine behave differently from sine between \(0^\circ\) and \(360^\circ\)?

**Tip:** Always remember that in the second quadrant, sine is positive, but cosine and tangent are negative.

Given that sin(x) = 5/13 where 90° ≤ x ≤ 180°, find the value of cos(x).

Learn how to find cos(x) given sin(x) = 5/13 where 90° ≤ x ≤ 180° using trigonometric identities. This problem involves the Pythagorean identity and the behavior of trigonometric functions in different quadrants. Suitable for students in Grades 9-12.

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