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 \( B C=5 \sqrt{3} \)

We are given two triangles to solve. ────────────────────────────── Example (a): Given:  Angle B = 60°, Angle A = 30°, side c = 40 Step 1. Find the third angle:  Angle C = 180° – (Angle A + Angle B)       = 180° – (30° + 60°)       = 90° Step 2. Use the Law of Sines:  a/sin A = b/sin B = c/sin C Since sin 30° = 1/2, sin 60° = √3/2, and sin 90° = 1, we have:  Common ratio = c/sin C = 40/1 = 40 Now compute:  side a = 40 × sin 30° = 40 × (1/2) = 20  side b = 40 × sin 60° = 40 × (√3/2) = 20√3 Thus, triangle (a) is completely solved:  Angle A = 30°, Angle B = 60°, Angle C = 90°  side a = 20, side b = 20√3, side c = 40 ────────────────────────────── Example (b): Given:  Angle A = 60°, Angle B = 30°, side BC = 5√3 Note: In triangle ABC the sides are usually assigned as follows:  side a = BC (opposite Angle A)  side b = AC (opposite Angle B)  side c = AB (opposite Angle C) Step 1. Find the third angle:  Angle C = 180° – (Angle A + Angle B)       = 180° – (60° + 30°)       = 90° Step 2. Use the Law of Sines:  a/sin A = b/sin B = c/sin C Here, side a = 5√3 and sin 60° = √3/2, sin 30° = 1/2, sin 90° = 1:  Common ratio = a/sin A = (5√3) / (√3/2) = 5√3 × (2/√3) = 10 Now compute:  side b = 10 × sin 30° = 10 × (1/2) = 5  side c = 10 × sin 90° = 10 × 1 = 10 Thus, triangle (b) is completely solved:  Angle A = 60°, Angle B = 30°, Angle C = 90°  side a = 5√3, side b = 5, side c = 10 ────────────────────────────── Final Answers: (a) Triangle ABC with Angle A = 30°, Angle B = 60°, Angle C = 90°:  side a = 20, side b = 20√3, side c = 40 (b) Triangle ABC with Angle A = 60°, Angle B = 30°, Angle C = 90°:  side a = 5√3, side b = 5, side c = 10