Angle bisectors

$AP$ is the angle bisector of $\angle A$ , $BP$ is the angle bisector of $\angle B$ . And $AP$ and $BP$ meets at $P$ . So $CP$ is also the angle bisector of $\angle C$

Now, $\angle C = 180^\circ - \angle A - \angle B$

$\Rightarrow \angle C = 70$

$\Rightarrow \angle PCA = 35$

$\Rightarrow \angle APC = 180^\circ - 22^\circ - 35^\circ$

$\Rightarrow \angle APC = \boxed {\boxed {123}}$

Note that P is the incenter of $\triangle ABC$ . (since it is the intersection of the two angle bisectors.) Thus, CP is the angle bisector of $\angle ACB$ . So, $\angle ACP = 35^{\circ}$ . Thus, $\angle APC = 123^{\circ}$ .

Since the sum of the angles of a triangles are 180, we have that $\angle BCA = 180 - 44 - 33 = 70$ . We also know that the three angle bisectors are concurrent, which means that since $P$ is on two of them, it must also be on the third and $\angle PCA = \dfrac{70}{2} = 35$ . Finally since the sum of the angles of $\triangle APC$ is 180, we can find that $\angle APC = 180 - 22 - 35 = \boxed{123^\circ}$

From the given information, we can deduce that $P$ is on the angle bisector of $\angle ABC$ and on the angle bisector of $\angle BAC$ . We know that the only point for which this happens is the incenter, and so $P$ is also on the angle bisector of $\angle ACB$ .

We can very easily find that $m \angle ACB = 70^\circ$ , which means that $m \angle ACP = 35^\circ$ . We use this and the fact that $m \angle PAC = 22^\circ$ to compute the desired value, $m \angle APC$ .

$m \angle APC = 180^\circ - 22^\circ - 35^\circ = \fbox{123}^\circ$ .

Since two angle bisectors determine a unique incenter, $P$ is the incenter of triangle $ABC$ . $\angle PCA=\frac{1}{2}(180^\circ-66^\circ-44^\circ)=35^\circ$ . $\angle APC=180^\circ-22^\circ-35^\circ=123^\circ$ .

(180-2(22)-2(33))/2=35

180-22-35=123

Make line AP and BP. Since AP bisect $\angle BAC$ and BP bisect $\angle ABC$ then AP and BP are angle bisectors, and we deduce that P is incenter of triangle ABC, So, CP is angle bisector too. We can easily get $\angle ACB = 70$ , $\angle ACP = 35$ and $\angle APC = 123$

Since $\angle PAB = \angle PAC$ and $\angle PBA = \angle PBC$ , $P$ is the incenter of $ABC$ . Thus, $\angle PCA = \angle PCB = \frac{180^\circ-2(22^\circ+33^\circ)}{2}=35^ \circ$ . Therefore, $\angle APC = 180^\circ - \angle PAC - \angle PCA = 180^\circ-22^\circ-35^\circ = 123^\circ$ .

Remark. We can find $\angle APC$ by using the formula $\angle APC = 90^\circ + \frac{ABC}{2} = 90^\circ + \angle PBC = 123^\circ$ . Note that this formula is obtained by similar arguments as above.