There are two possible triangles with $AB = 13,$ $BC = 10,$ and $A = 40^\circ.$  One of them is shown below.  What is the sum of the two possible values of $\angle B,$ in degrees?

[asy]
unitsize (1 cm);

pair A, B, C;

A = (0,0);
B = 5*dir(40);
C = (5,0);

draw(A--B--C--cycle);

label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$13$", (A + B)/2, NW);
label("$10$", (B + C)/2, NE);
label("$40^\circ$", (1,0.4));
[/asy]
Let the two possible positions of $C$ be $C_1$ and $C_2,$ as shown below.  Then the two possible values of $\angle B$ are $\angle ABC_1$ and $\angle ABC_2.$

[asy]
unitsize (1 cm);

pair A, B;
pair[] C;

A = (0,0);
B = 5*dir(40);
C[1] = (2*B.x - 5,0);
C[2] = (5,0);

draw(A--B--C[2]--cycle);
draw(B--C[1]);

label("$A$", A, SW);
label("$B$", B, N);
label("$C_1$", C[1], S);
label("$C_2$", C[2], SE);
label("$13$", (A + B)/2, NW);
label("$10$", (B + C[2])/2, NE);
label("$10$", (B + C[1])/2, NW);
label("$40^\circ$", (1,0.4));
[/asy]

Note that
\[\angle ABC_1 = 180^\circ - 40^\circ - \angle AC_1 B = 140^\circ - \angle AC_1 B\]and
\[\angle ABC_2 = 180^\circ - 40^\circ - \angle AC_2 B = 140^\circ - \angle AC_2 B.\]Since $\angle AC_1 B = 180^\circ - \angle BC_1 C_2 = 180^\circ - \angle AC_2 B,$
\begin{align*}
\angle ABC_1 + \angle ABC_2 &= (140^\circ - \angle AC_1 B) + (140^\circ - \angle AC_2 B) \\
&= 280^\circ - (\angle AC_1 B + \angle AC_2 B) \\
&= 280^\circ - 180^\circ = \boxed{100^\circ}.
\end{align*}