Point D on side AB of triangle ABC is chosen so that AD = AC. It is known that ∠CAB = 122 ° and ∠ACB = 47 °. 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 ° = 122 ° + ∠ADC + ∠ACD
∠ADC + ∠ACD = 58 °
Since AD = AC, this triangle is isosceles.
Then, ∠ADC = ∠ACD (by the property of an isosceles triangle), we obtain that:
∠ADC = ∠ACD = 58 ° / 2 = 29 °
∠DCB = ∠ACB-∠ACD = 47 ° -29 ° = 18 °
Answer: 18

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