Exactly three of the interior angles of a convex [polygon](https://artofproblemsolving.com/wiki/index.php/Polygon) are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \  } 6 \qquad \mathrm{(D) \  } 7 \qquad \mathrm{(E) \  }8$
All angle measures are in degrees.
The sum of the interior angle measures of an $n$-gon is $180(n-2)=180n-360$. Let the three obtuse angle measures be $o_1, o_2,$ and $o_3$, and the $n-3$ acute angle measures be $a_1, a_2, a_3, \cdots$. 

Since $90<o_i<180$, $3(90)=270<o_1+o_2+o_3<3(180)=540$. 

Similarly, since $0<a_i<90$, $0(n-3)=0<a_1+a_2+a_3+\cdots<90(n-3)=90n-270$. 

Thus, $270+0<o_1+o_2+o_3+a_1+a_2+a_3+\cdots<540+90n-270$ $\implies 270<180n-360<90n+270$.

Thus, $180n-360<90n+270\implies n<7$, so the largest possible $n$ is $6, \boxed{\text{C}}$.