Find angle in two triangles

$constuct\quad a\quad point\quad E\quad where\quad AE=CB,EC=AB\quad and\quad AE//BC,EC//AB\\ \Rightarrow AECB\quad is\quad //gram.\Rightarrow \angle ABC=AEC=50°\\ \angle ADC=180°-20°-30°=130°\\ \angle ADC+\angle AEC=180°\Rightarrow ADCE\quad is\quad concyclic\\ \because AD=BC=AE\Rightarrow ACD=ACE=30°\Rightarrow EAC=180°-30°-50°=100°\\ ACB=EAC=100°\Rightarrow DCB=100°-30°=70°$

Applying the law of sines to the triangle ACD, we obtain

(1) $\frac{AD}{\sin(\angle DCA)} = \frac{AC}{\sin(\angle ADC)}$ ,

or, since $\angle DCA = 30^{\circ}$ and $\angle ADC = 180^{\circ} - \angle DAC - \angle DCA = 130^{\circ}$ ,

(2) $\frac{AD}{AC} = \frac{\sin 30^{\circ}}{\sin 130^{\circ}}$ .

Analogously, in the triangle ABC we have

(3) $\frac{BC}{\sin(\angle BAC)} = \frac{AC}{\sin(\angle ABC)}$ ,

or, since $\angle ABC = 50^{\circ}$ and $\angle BAC = 180^{\circ} - \angle ABC - \angle ACB = 130^{\circ} - \angle ACB$ ,

(4) $\frac{BC}{AC} = \frac{\sin(130^{\circ}-\angle ACB)}{\sin 50^{\circ}}$ .

Since $AD = BC$ , it follows from (2) and (4) that

(5) $\frac{\sin 30^{\circ}}{\sin 130^{\circ}} = \frac{\sin(130^{\circ}-\angle ACB)}{\sin 50^{\circ}}$ .

One can simplify (5) by noticing that $\sin 130^{\circ} = \sin 50^{\circ}$ , so that $\sin 30^{\circ} = \sin(130^{\circ}-\angle ACB)$ , from which follows $\angle ACB = 100^{\circ}$ . Finally, we obtain the desired result by subtracting $\angle DCA$ from $\angle ACB$ :

(6) $\angle DCB = \angle ACB - \angle DCA = 100^{\circ} - 30^{\circ} = 70^{\circ}. \,\square$