1. **Problem statement:** Find the angle $A$ in the third quadrant such that $\cos A = \sin \left( \frac{5\pi}{3} \right)$. 2. **Recall the sine value:** Calculate $\sin \left( \frac{5\pi}{3} \right)$. Since $\frac{5\pi}{3} = 2\pi - \frac{\pi}{3}$, this angle lies in the fourth quadrant where sine is negative. The reference angle is $\frac{\pi}{3}$. Thus, $$\sin \left( \frac{5\pi}{3} \right) = -\sin \left( \frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}.$$ 3. **Express the condition:** We want $$\cos A = -\frac{\sqrt{3}}{2}.$$ 4. **Determine angles where $\cos A = -\frac{\sqrt{3}}{2}$:** Cosine is negative in the second and third quadrants. The reference angle for $\cos \theta = \frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$. Hence the angles with $\cos A = -\frac{\sqrt{3}}{2}$ are $$A = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$ (second quadrant), and $$A = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$$ (third quadrant). 5. **Choose the angle in the third quadrant:** Since the question asks for an angle in the third quadrant, the correct choice is $$\boxed{\frac{7\pi}{6}}.$$ --- **Final answer:** $A = \frac{7\pi}{6}$ is the angle in the third quadrant where $\cos A = \sin \left( \frac{5\pi}{3} \right)$.