Triangle ABC Angle A is 45 degrees Angle C is 15 degrees BC is 4√6 Find: AB AC and angle B.

Two angles of the triangle are known by condition, we find the third:
∠ B = 180 ° – (∠ A + ∠ C) = 180 ° – 60 ° = 120 °.
To find the sides, we use the sine theorem.
BC / sin A = AC / sin B = 4√6 / √2 / 2 = AC / √3 / 2 →
AC = 4√6 * √3 / 2 * 2 / √2 = 4√18 / √2 = 12.
BC / sin A = AB / sin C →
AB = BC * sin 15 ° / sin 45 ° = 4√6 * 0.258819 / √2 / 2 = 3.58633 ≈ 3.6.
Answer: Angle B is 120 °, side AB is 3.6, side AC is 12.