In triangle ABC, angle A = 60 degrees. AC / AB = 1/2. Find the angle B in degrees.

Let’s designate the side | AC | through a, then | AB | = 2a, using the cosine theorem we obtain:

| BC | ^ 2 = a ^ 2 + (2a) ^ 2 – 2 * a * 2a * cos (60) = 5a ^ 2 – 2a ^ 2 = 3a ^ 2;

| BC | = √3 * a.

By the sine theorem, we get:

| BC | / sin (A) = | AC | / sin (B);

sin (B) = a * sin (60) / √3 * a = 1/2.

B = arcsin (1/2) = 30 degrees.