4. In $\triangle A B C, \angle A C B=36^{\circ}$ and the interior angle bisectors of $\angle C A B$ and $\angle A B C$ intersect at $P$. Find $\angle A P B$.
(A) $72^{\circ}$
(B) $108^{\circ}$
(C) $126^{\circ}$
(D) $136^{\circ}$
(E) None of the above
4. Answer: (B)
Let $\angle C A B=x$ and $\angle A B C=y$. Then $x+y=180^{\circ}-36^{\circ}=144^{\circ}$.
Now $\angle A P B=180^{\circ}-\frac{x+y}{2}=108^{\circ}$.