14. If a convex polygon has exactly three interior angles that are obtuse, what is the maximum number of sides this polygon can have?
(A) 4;
(B) 5 ;
(C) 6 ;
(D) 7 ;
(E) 8 .
For $(n-2) \cdot 180^{\circ}$. If the polygon has exactly 3 obtuse angles, then the remaining $n-3$ angles are all no more than $90^{\circ}$. Therefore,
$$
(n-2) 180^{\circ}<(n-3) 90^{\circ}+3 \cdot 180^{\circ} \text {, }
$$

i.e., $2(n-2)<n-3+6, n<7$.
The following figure shows:

$n=6$ is possible
(Here, $\triangle A B C$ is an equilateral
triangle, $\triangle A P B$, $\triangle B$
$Q C$,
$\triangle A R C$ are all isosceles
triangles with an angle of
$160^{\circ}$.
(C) is true.