Given the conditions in the triangle $ABC$ with point $P$ inside the triangle, we need to find the measure of angle $\angle PBA$.

Problem Analysis:

• $\angle PAC = 40^\circ$
• $\angle PAB = 30^\circ$
• $\angle PCA = 20^\circ$
• $\angle PCB = 10^\circ$

Solution:

Let $x = \angle PBA$, which is what we need to find.

Step 1: Use the sum of angles around point $P$

The angles $\angle PAC$, $\angle PAB$, $\angle PCA$, and $\angle PCB$ are connected to the angles at point $P$. The sum of angles around point $P$ inside the triangle is $360^\circ$. This leads us to consider the sum of angles in smaller parts of triangle ABC that include point P.

Step 2: Finding angle $\angle APC$

$\angle APC = 180^\circ - (\angle PAC + \angle PCA) = 180^\circ - (40^\circ + 20^\circ) = 120^\circ$

Step 3: Finding angle $\angle APB$

$\angle APB = 180^\circ - (\angle PAB + \angle PBA) = 180^\circ - (30^\circ + x)$

Step 4: Using triangle sum for triangle PBC

In triangle $PBC$, the sum of internal angles is:

$\angle PBC + \angle PCB + \angle APB = 180^\circ$

Substituting the known values:

$x + 10^\circ + (150^\circ - x) = 180^\circ$

Simplify the equation:

$120^\circ = x$

Hence the measure is x = 120^\circ.

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Did you have any clarifications on that, questions for further answerings?