由$$\angle A+\angle B+\angle C=180{}^{}\circ $$，即$$\left( \theta -\alpha \right)+\theta +\left( \theta +\alpha \right)=3\theta =180{}^{}\circ $$，  所以$$\theta =\angle ABC=60{}^{}\circ $$，  因此$$\frac{1}{2}\left( \angle BAC+\angle BCA \right)=\frac{1}{2}\left( 180{}^{}\circ -60{}^{}\circ \right)=60{}^{}\circ $$，  所以$$\angle APC=180{}^{}\circ -\frac{1}{2}\left( \angle BAC+\angle BCA \right)=180{}^{}\circ -60{}^{}\circ =120{}^{}\circ $$．  故应选$$\text{C}$$． 
在$$\triangle ABC$$中，$$\angle A=\theta -\alpha $$，$$\angle B=\theta $$，$$\angle C=\theta +\alpha $$，（$$0{}^{}\circ \textless{}\alpha \textless{}\theta \textless{}90{}^{}\circ $$） ．若$$\angle BAC$$与$$\angle BCA$$的平分线相交于$$P$$点，则$$\angle APC$$等于．
Options:
1: $$90{}^{}\circ $$ 
2: $$105{}^{}\circ $$ 
3: $$120{}^{}\circ $$ 
4: $$150{}^{}\circ $$ 