Given a convex quadrilateral $ABCD$ in which $\angle BAC = 20^o$,  $\angle CAD = 60^o$,  $\angle ADB = 50^o$ , and $\angle BDC = 10^o$. Find $\angle ACB$.
1. **Identify the given angles and the structure of the quadrilateral:**
   - $\angle BAC = 20^\circ$
   - $\angle CAD = 60^\circ$
   - $\angle ADB = 50^\circ$
   - $\angle BDC = 10^\circ$

2. **Analyze $\triangle ACD$:**
   - Since $\angle CAD = 60^\circ$, we need to check if $\triangle ACD$ is an equilateral triangle.
   - If $\triangle ACD$ is equilateral, then $\angle ACD = 60^\circ$ and $\angle ADC = 60^\circ$.

3. **Calculate $\angle ABD$:**
   - $\angle ABD = 180^\circ - \angle BAC - \angle CAD - \angle ADB$
   \[
   \angle ABD = 180^\circ - 20^\circ - 60^\circ - 50^\circ = 50^\circ
   \]

4. **Determine the lengths of sides:**
   - Since $\angle ABD = 50^\circ$ and $\angle ADB = 50^\circ$, $\triangle ABD$ is isosceles with $AB = AD$.
   - Given $\angle CAD = 60^\circ$, if $\triangle ACD$ is equilateral, then $AC = AD$.

5. **Conclude that $\triangle BAC$ is isosceles:**
   - Since $AB = AD$ and $AC = AD$, it follows that $AB = AC$.
   - Therefore, $\triangle BAC$ is isosceles with $\angle BAC = 20^\circ$.

6. **Calculate $\angle ACB$:**
   - In $\triangle BAC$, since it is isosceles with $AB = AC$, the base angles are equal.
   - Let $\angle ACB = \angle ABC = x$.
   - The sum of angles in $\triangle BAC$ is $180^\circ$:
   \[
   \angle BAC + \angle ACB + \angle ABC = 180^\circ
   \]
   \[
   20^\circ + x + x = 180^\circ
   \]
   \[
   20^\circ + 2x = 180^\circ
   \]
   \[
   2x = 160^\circ
   \]
   \[
   x = 80^\circ
   \]

Therefore, $\angle ACB = 80^\circ$.

The final answer is $\boxed{80^\circ}$