1. **State the problem:** We have a right triangle with \(\angle C = 90^\circ\), \(\angle A = 30^\circ\), and \(\angle B = 60^\circ\). The side opposite \(\angle B\) (side AC) is 18 units. We need to find the lengths of sides \(a\) and \(c\).\n\n2. **Recall the triangle angle sum rule:** The sum of angles in a triangle is \(180^\circ\). Since \(\angle C = 90^\circ\) and \(\angle A = 30^\circ\), then \(\angle B = 180^\circ - 90^\circ - 30^\circ = 60^\circ\) (given).\n\n3. **Use trigonometric ratios:** In a right triangle, the sides relate to angles as follows:\n- Opposite side to \(\theta\): side opposite the angle\n- Adjacent side to \(\theta\): side next to the angle (not hypotenuse)\n- Hypotenuse: longest side opposite the right angle\n\n4. **Label sides:**\n- Side \(a\) is opposite \(\angle A = 30^\circ\)\n- Side \(b\) is opposite \(\angle B = 60^\circ\) (given as 18 units)\n- Side \(c\) is the hypotenuse\n\n5. **Use sine and cosine for \(\angle B = 60^\circ\):**\n\n\[ \sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c} \Rightarrow \sin 60^\circ = \frac{18}{c} \]\n\n6. **Solve for \(c\):**\n\[ c = \frac{18}{\sin 60^\circ} \]\n\nSince \(\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866\),\n\[ c = \frac{18}{0.866} \approx 20.8 \]\n\n7. **Use cosine for \(\angle B = 60^\circ\) to find side \(a\):**\n\[ \cos 60^\circ = \frac{a}{c} \Rightarrow a = c \times \cos 60^\circ \]\n\nSince \(\cos 60^\circ = 0.5\),\n\[ a = 20.8 \times 0.5 = 10.4 \]\n\n**Final answers:**\n\[ m\angle B = 60^\circ, \quad a \approx 10.4 \text{ units}, \quad c \approx 20.8 \text{ units} \]