$$ABCD$$ is a quadrilateral in which $$\angle BAC = 20^\circ$$, $$\angle CAD = 60^\circ$$, $$\angle ADB = 50^\circ$$, $$\angle BDC = 10^\circ$$. Then $$\angle ACB = $$

In $$\triangle ABD$$, $$\angle BAD = 20^\circ + 60^\circ = 80^\circ$$ and $$\angle ADB = 50^\circ$$, so $$\angle ABD = 50^\circ$$ and hence $$AB = AD$$. In $$\triangle ACD$$, $$\angle CAD = 60^\circ$$ and $$\angle ADC = 50^\circ + 10^\circ = 60^\circ$$, so the triangle is equilateral and $$AC = AD$$. Therefore $$AB = AC$$, and in the isosceles triangle $$ABC$$ with $$\angle BAC = 20^\circ$$ we get $$\angle ACB = \frac{180^\circ - 20^\circ}{2} = 80^\circ$$.