Question

Point X is on side AC of triangle ABC such that <AXB =<ABX, and <ABC - <ACB = 39 degrees. Find <XBC in degrees.

Answer 1

This is hard to solve if you don't draw it. So I hope you'll draw the figure, and I'll provide the steps here. Anyway, first, I'll rename some angles.
Let angle AXB = n
Let angle ABX = m
Let angle ABC = B
Let angle ACB = C
It was said in the problem that angles AXB and ABX are equal:
m = n : equation (1)

Also, it was stated in the problem the relationship between angles ABC and ACB:
B - C = 39
B = 39 + C : equation (2)

Figure ABX is a triangle. Thus the sum of its interior angles is equal to 180:
A + m + n = 180
Substituting equation (1) here,
A + 2m = 180
A = 180 - 2m : equation (3)

Figure ABC is a triangle, thus the sum of its interior angles is equal to 180:
A + B + C = 180
Substituting equations (2) and (3) here,
180 - 2m + 39 + C + C = 180
-2m + 2C + 39 = 0
2m = 39 + 2C
m = 19.5 + C : equation (4)

Figure XBC is a triangle, and angle BXC is the supplementary of angle m. Thus,
BXC = 180 - m : equation (5)
BXC + XBC + C = 180
We're solving for angle XBC. Substituting equations (4) and (5) here,
180 - (19.5 + C) + XBC + C = 180
-19.5 - C + XBC + C = 0
XBC = 19.5 degrees

hope this helps~ `u`