1. The problem is to find the angle $A$ such that $\cos A = \sin \left( \frac{5\pi}{3} \right)$.\n\n2. Recall the identity $\sin x = \cos \left( \frac{\pi}{2} - x \right)$. Applying this gives $\sin \left( \frac{5\pi}{3} \right) = \cos \left( \frac{\pi}{2} - \frac{5\pi}{3} \right)$.\n\n3. Calculate inside the cosine: $\frac{\pi}{2} - \frac{5\pi}{3} = \frac{3\pi}{6} - \frac{10\pi}{6} = -\frac{7\pi}{6}$.\n\n4. So, $\sin \left( \frac{5\pi}{3} \right) = \cos \left( -\frac{7\pi}{6} \right)$.\n\n5. Since cosine is an even function, $\cos (-x) = \cos x$, so $\cos \left( -\frac{7\pi}{6} \right) = \cos \left( \frac{7\pi}{6} \right)$.\n\n6. Therefore, $\cos A = \cos \left( \frac{7\pi}{6} \right)$, which means $A = \pm \frac{7\pi}{6} + 2k\pi$ for any integer $k$.\n\n7. From the given options, Option 4: $A = \frac{7\pi}{6}$ satisfies this.\n\nFinal answer: $A = \frac{7\pi}{6}$ (Option 4).