B. In a convex $n$-sided polygon, what is the maximum number of interior angles that can be $150^{\circ}$? Explain your reasoning.
B. Suppose in a convex $n$-sided polygon, there are $k$ interior angles equal to $150^{\circ}$. Then there are $n-k$ interior angles not equal to $150^{\circ}$.
(1) If $k=n$, then
$$
n \times 150^{\circ}=(n-2) \times 180^{\circ},
$$

we get $n=12$. Thus, in a regular dodecagon, all 12 interior angles are $150^{\circ}$.
(2) If $k(n-2) \times 180^{\circ}$,
$$
we get $k<12$, i.e., $k \leqslant 11$.

When $k=11$, there exists a convex $n$-sided polygon with 11 interior angles of $150^{\circ}$, and the remaining $n-k$ angles are
$$
\begin{aligned}
\alpha & =\frac{(n-2) \times 180^{\circ}-11 \times 150^{\circ}}{n-11} \\
& =\left(6-\frac{1}{n-11}\right) \times 30^{\circ} \in\left[165^{\circ}, 180^{\circ}\right) .
\end{aligned}
$$
(3) If $k<n$, and $8 \leqslant n \leqslant 11$, when $k=n-1$, there exists a convex $n$-sided polygon with $n-1$ interior angles of $150^{\circ}$, and the other angle is
$$
\begin{array}{l}
\alpha=(n-2) \times 180^{\circ}-(n-1) \times 150^{\circ} \\
=(n-7) \times 30^{\circ} \in\left[30^{\circ}, 120^{\circ}\right] .
\end{array}
$$
(4) If $k<n$, and $3 \leqslant n \leqslant 7$, from (3) we know $k \leqslant n-2$, when $k=n-2$, there exists a convex $n$-sided polygon with $n-2$ interior angles of $150^{\circ}$, and the other two angles are both $(n-2) \times 15^{\circ}$.
In summary, when $n=12$, the maximum value of $k$ is 12;
when $n \geqslant 13$, the maximum value of $k$ is 11;
when $8 \leqslant n \leqslant 11$, the maximum value of $k$ is $n-1$;
when $3 \leqslant n \leqslant 7$, the maximum value of $k$ is $n-2$.