1. The problem asks which expressions have the same value as $\cos 50^\circ$.\n\n2. Recall the co-function identity: $\cos \theta = \sin (90^\circ - \theta)$. So, $\cos 50^\circ = \sin 40^\circ$.\n\n3. Also, cosine and sine have periodic and symmetry properties:\n- $\sin (180^\circ - x) = \sin x$\n- $\cos (360^\circ - x) = \cos x$\n- $\cos (180^\circ - x) = -\cos x$\n\n4. Evaluate each option:\nA. $\sin 40^\circ = \cos 50^\circ$ (by co-function identity)\nB. $\sin 130^\circ = \sin (180^\circ - 50^\circ) = \sin 50^\circ \neq \cos 50^\circ$\nC. $\cos 130^\circ = \cos (180^\circ - 50^\circ) = -\cos 50^\circ \neq \cos 50^\circ$\nD. $\cos 140^\circ = \cos (180^\circ - 40^\circ) = -\cos 40^\circ \neq \cos 50^\circ$\nE. $\sin 220^\circ = \sin (180^\circ + 40^\circ) = -\sin 40^\circ \neq \cos 50^\circ$\nF. $\cos 230^\circ = \cos (180^\circ + 50^\circ) = -\cos 50^\circ \neq \cos 50^\circ$\nG. $\cos 310^\circ = \cos (360^\circ - 50^\circ) = \cos 50^\circ$\nH. $\sin 320^\circ = \sin (360^\circ - 40^\circ) = -\sin 40^\circ \neq \cos 50^\circ$\n\n5. Therefore, the expressions equal to $\cos 50^\circ$ are A and G.\n\nFinal answer: A and G.