The angles $A$ and $B$ of base of the isosceles triangle $ABC$ are equal to $40^o$. Inside $\vartriangle ABC$, $P$ is such that $\angle PAB = 30^o$ and $\angle PBA = 20^o$. Calculate, without table, $\angle PCA$.
1. **Identify the given angles and properties of the triangle:**
   - The triangle \( \triangle ABC \) is isosceles with \( \angle A = \angle B = 40^\circ \).
   - Inside \( \triangle ABC \), point \( P \) is such that \( \angle PAB = 30^\circ \) and \( \angle PBA = 20^\circ \).

2. **Calculate the remaining angles:**
   - Since \( \angle A + \angle B + \angle C = 180^\circ \) in any triangle, we have:
     \[
     \angle C = 180^\circ - \angle A - \angle B = 180^\circ - 40^\circ - 40^\circ = 100^\circ
     \]

3. **Determine \( \angle APB \):**
   - Using the given angles \( \angle PAB = 30^\circ \) and \( \angle PBA = 20^\circ \):
     \[
     \angle APB = 180^\circ - \angle PAB - \angle PBA = 180^\circ - 30^\circ - 20^\circ = 130^\circ
     \]

4. **Use the Law of Sines to find the ratio \( \frac{PB}{AB} \):**
   - By the Law of Sines in \( \triangle APB \):
     \[
     \frac{PB}{\sin \angle PAB} = \frac{AB}{\sin \angle APB}
     \]
     Substituting the known values:
     \[
     \frac{PB}{\sin 30^\circ} = \frac{AB}{\sin 130^\circ}
     \]
     Since \( \sin 30^\circ = \frac{1}{2} \) and \( \sin 130^\circ = \sin (180^\circ - 50^\circ) = \sin 50^\circ \):
     \[
     \frac{PB}{\frac{1}{2}} = \frac{AB}{\sin 50^\circ}
     \]
     \[
     PB = AB \cdot \frac{\sin 30^\circ}{\sin 130^\circ} = AB \cdot \frac{\frac{1}{2}}{\sin 50^\circ} = AB \cdot \frac{1}{2 \sin 50^\circ}
     \]

5. **Simplify the expression for \( PB \):**
   - Using the identity \( \sin 50^\circ = \cos 40^\circ \):
     \[
     PB = AB \cdot \frac{1}{2 \cos 40^\circ}
     \]
     Since \( \cos 40^\circ = \sin 50^\circ \):
     \[
     PB = AB \cdot \frac{\cos 50^\circ}{2 \sin 50^\circ \cos 50^\circ} = AB \cdot \frac{\cos 50^\circ}{\sin 100^\circ} = AB \cdot \frac{\sin 40^\circ}{\sin 100^\circ}
     \]
     \[
     PB = AB \cdot \frac{\sin \angle CAB}{\sin \angle ACB} = CB
     \]

6. **Calculate \( \angle CBP \) and \( \angle PCB \):**
   - Since \( \angle CBP = \angle CBA - \angle PBA = 40^\circ - 20^\circ = 20^\circ \):
     \[
     \angle PCB = 180^\circ - \angle CBP - \angle BPC = 180^\circ - 20^\circ - 80^\circ = 80^\circ
     \]

7. **Determine \( \angle PCA \):**
   - Since \( \angle PCA = \angle ACB - \angle PCB = 100^\circ - 80^\circ = 20^\circ \):
     \[
     \angle PCA = 20^\circ
     \]

The final answer is \( \boxed{20^\circ} \)