1985 AMC 12 Problem 14

Bound the three obtuse angles by $180^\circ$ and every other angle by $90^\circ$

Compare that upper bound with the interior-angle sum $(n-2)180^\circ$

For an $n$-gon, the three obtuse angles have total less than $3\cdot180^\circ,$ while the other $n-3$ angles have total at most $(n-3)90^\circ.$ Hence $$\begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ, \end{aligned}$$ which simplifies to $n\lt7.$ Six sides are attainable, for example with interior angles $150^\circ,$ $150^\circ,$ $150^\circ,$ $90^\circ,$ $90^\circ,$ and $90^\circ.$ Thus the maximum is $6.$

Therefore the correct answer is C.