Angle hunting

Since the angle sum of a triangle is $180°$ , $\angle APB = 135°$ , $\angle ARC = 120°$ , and $\angle BQC = 105°$ .

By the law of sines on $\triangle APB$ , $\triangle ARC$ , and $\triangle BQC$ , $AP = \frac{\sin 30°}{\sin 135°}AB = \frac{\sqrt{2}}{2}AB$ , $BP = \frac{\sin 15°}{\sin 135°}AB = \frac{\sqrt{3} - 1}{2}AB$ , $AR = \frac{\sin 45°}{\sin 120°}AC = \frac{\sqrt{6}}{3}AC$ , $CR = \frac{\sin 15°}{\sin 120°}AC = \frac{3\sqrt{2} - \sqrt{6}}{6}AC$ , $BQ = \frac{\sin 45°}{\sin 105°}BC = \frac{\sqrt{3} - 1}{2}BC$ , and $CQ = \frac{\sin 30°}{\sin 105°}BC = \frac{\sqrt{6} - \sqrt{2}}{2}BC$ .

By the law of cosines on $\triangle APR$ , $\triangle BPQ$ , and $\triangle CRQ$ ,

$PR^2 = AP^2 + AR^2 - 2 \cdot AP \cdot AR \cdot \cos(\angle BAC + 30°)$

$PQ^2 = BP^2 + BQ^2 - 2 \cdot BP \cdot BQ \cdot \cos(\angle ABC + 60°)$

$QR^2 = CQ^2 + CR^2 - 2 \cdot CQ \cdot CR \cdot \cos(\angle ACB + 90°)$

Substituting the above values of $AR$ , $AP$ , $BP$ , $BQ$ , $CQ$ , and $CR$ , using the the identity $\cos(a + b) = \cos a \cos b - \sin a \sin b$ , the law of cosine formulas from $\triangle ABC$ ( $\cos \angle BAC = \frac{AC^2 + AB^2 - BC^2}{2 \cdot AC \cdot AB}$ , $\cos \angle ABC = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC}$ , and $\cos \angle ACB = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC}$ ), and the area $K$ formula from $\triangle ABC$ ( $K = \frac{1}{2}AC \cdot BC \sin \angle ACB$ $=$ $\frac{1}{2}AB \cdot AC \sin \angle BAC$ $=$ $\frac{1}{2}AB \cdot BC \sin \angle ABC$ ), $PR^2$ , $PQ^2$ , and $QR^2$ simplify to:

$PR^2 = \frac{1}{2}BC^2 + \frac{1}{6}AC^2 + \frac{2\sqrt{3}}{3}K$

$PQ^2 = (3 - \frac{3\sqrt{3}}{2})BC^2 + (1 - \frac{\sqrt{3}}{2})AC^2 + (4\sqrt{3} - 6)K$

$QR^2 = (2 - \sqrt{3})BC^2 + (\frac{2}{3} - \frac{\sqrt{3}}{3})AC^2 + (\frac{8\sqrt{3}}{3} - 4)K$

Then by the law of cosines on $\triangle PQR$ , $\cos \angle PRQ = \frac{PR^2 + QR^2 - PQ^2}{2 \cdot PR \cdot QR}$ which simplifies to $\cos \angle PRQ = \frac{(\frac{1}{2} + \frac{\sqrt{3}}{2})BC^2 + (-\frac{1}{6} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2})AC^2 + (2 - \frac{2\sqrt{3}}{3})K}{2\sqrt{(\frac{1}{2} + \frac{\sqrt{3}}{2})BC^2 + (-\frac{1}{6} - \frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2})AC^2 + (2 - \frac{2\sqrt{3}}{3})K}^2}$ , which reduces further to $\cos \angle PRQ = \frac{1}{2}$ , meaning $\angle PRQ = 60°$ (regardless of $\triangle ABC$ !)

Similarly, $\cos \angle QPR = \frac{PR^2 + PQ^2 - QR^2}{2 \cdot PR \cdot PQ}$ which simplifies to $\cos \angle QPR = \frac{\sqrt{2}((\frac{3\sqrt{2}}{4} + \frac{\sqrt{6}}{4})BC^2 + (\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{12})AC^2 + (\sqrt{6} - \sqrt{2})K)}{2\sqrt{(\frac{3\sqrt{2}}{4} + \frac{\sqrt{6}}{4})BC^2 + (\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{12})AC^2 + (\sqrt{6} - \sqrt{2})K}^2}$ , which reduces further to $\cos \angle PRQ = \frac{\sqrt{2}}{2}$ , meaning $\angle PRQ = 45°$ (also regardless of $\triangle ABC$ ).

Finally, since the angle sum of a triangle is $180°$ , and since $\angle PRQ = 45°$ and $\angle PRQ = 60°$ , $\angle PRQ = 180° - 45° - 60° = 75°$ .

So regardless of $\triangle ABC$ , the three angles in $\triangle PQR$ are always $\angle PRQ = 45°$ , $\angle PRQ = 60°$ , and $\angle PRQ = 75°$ , and the largest of these angles is $\boxed{75}$ .