3. Solve \( \triangle \mathrm{ABC} \) if (a) \( \angle \mathrm{B}=60^{\circ}, \angle \mathrm{A}=30^{\circ} \) and \( \mathrm{c}=40 \) (b) \( \angle \mathrm{A}=60^{\circ}, \angle B=30^{\circ} \) and \( B C=5 \sqrt{3} \)

To solve the triangles, we can employ the Law of Sines and other trigonometric principles. For part (a): Given \( \angle B = 60^\circ \), \( \angle A = 30^\circ \), and \( c = 40 \), we first find \( \angle C \): \[ \angle C = 180^\circ - \angle A - \angle B = 180^\circ - 30^\circ - 60^\circ = 90^\circ. \] Now we can find \( a \) and \( b \) using the Law of Sines: \[ \frac{a}{\sin A} = \frac{c}{\sin C}, \quad \frac{b}{\sin B} = \frac{c}{\sin C}. \] Thus: \[ \frac{a}{\sin 30^\circ} = \frac{40}{\sin 90^\circ} \implies a = 40 \cdot \frac{1}{2} = 20, \] \[ \frac{b}{\sin 60^\circ} = \frac{40}{\sin 90^\circ} \implies b = 40 \cdot \frac{\sqrt{3}}{2} \approx 34.64. \] So, the sides are \( a = 20 \), \( b \approx 34.64 \), and \( c = 40 \). For part (b): Given \( \angle A = 60^\circ \), \( \angle B = 30^\circ \), and \( BC = 5\sqrt{3} \). First, calculate \( \angle C \): \[ \angle C = 180^\circ - 60^\circ - 30^\circ = 90^\circ. \] Then using the Law of Sines: \[ \frac{a}{\sin A} = \frac{BC}{\sin C} \implies a = 5\sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{15}{2} = 7.5, \] and \[ \frac{b}{\sin B} = \frac{BC}{\sin C} \implies b = 5\sqrt{3} \cdot \frac{1}{2} = \frac{5\sqrt{3}}{2} \approx 4.33. \] So, the sides are \( a = 7.5 \), \( b \approx 4.33 \), and \( BC = 5\sqrt{3} \). Happy triangulation!