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.