5) The answer is (B)

Indeed, remembering that the sum of the interior angles of a triangle is $180^{\circ}$ and that $\mathrm{AN}$ and $\mathrm{CM}$ are the angle bisectors of the angles at $\mathrm{A}$ and $\mathrm{C}$, we have that

$$
\hat{B}=180^{\circ}-\hat{A}-\hat{C}=180^{\circ}-2(P \hat{A} C+P \hat{C A})=180^{\circ}-2\left(180^{\circ}-A \hat{P C}\right)=100^{\circ} \text {. }
$$
5) In triangle $A B C$, the rays $A N$ and $C M$ are the bisectors of $B \widehat{A} C$ and $B \widehat{C} A$ and intersect at $P$. Knowing that $A \widehat{P} C=140^{\circ}$, what is the measure of the angle at $B$?
(A) $90^{\circ}$
(B) $100^{\circ}$
(C) $110^{\circ}$
(D) $120^{\circ}$
(E) $130^{\circ}$