Point D on side AB of triangle ABC is chosen so that AD = AC. It is known that ∠CAB = 19 ° and ∠ACB = 160 °. Find the angle of the DCB.

Consider the triangle ACD.
By the theorem on the sum of the angles of a triangle:
180 ° = ∠CAB + ∠ADC + ∠ACD
180 ° = 19 ° + ∠ADC + ∠ACD
∠ADC + ∠ACD = 161 °
Since AD = AC, this triangle is isosceles.
Then, ∠ADC = ∠ACD (by the property of an isosceles triangle), we obtain that:
∠ADC = ∠ACD = 161 ° / 2 = 80.5 °
∠DCB = ∠ACB-∠ACD = 160 ° -80.5 ° = 79.5 °
Answer: 79.5