A side and two angles of a triangle ABC are given. Find its third corner and the other two sides

A side and two angles of a triangle ABC are given. Find its third corner and the other two sides if the angle B = 40; BC = 3.5; angle C = 120 degrees.

Let us find out what the angle A will equal if from the condition of the task we know that the angle B corresponds to 40 °, while the angle C is equal to 120 °:
180 – 40 – 120 = 20.
Now it’s time to apply the theorem of sines known to us from the school curriculum, according to which, the ratios of the sines of the angles to opposite sides are equal.
Let’s define what side AB will be equal to:
sin 20 ° / BC = sin 120 ° / AB;
AB = BC * sin 120 ° / sin 20 ° = 3.5 * √3 / 2: 0.342 ≈ 5.12√3.
Then the AC side:
sin 20 ° / BC = sin 40 ° / AC;
AC = 3.5 * 0.6428: 0.342 ≈ 6.58.
Answer: ∡A = 20 °, AB = 5.12√3, AC = 6.58.