$19 \cdot 30$ If a convex polygon has exactly three obtuse interior angles, the maximum number of sides this polygon can have is
(A) 4 .
(B) 5 .
(C) 6 .
(D) 7 .
(E) 8 .
(36th American High School Mathematics Examination, 1985)
[Solution 1] Since the sum of the exterior angles of a convex polygon is $360^{\circ}$, the number of non-acute exterior angles is no more than 3 (and given that it has three obtuse angles, there are already three acute angles among the exterior angles), which means the number of non-obtuse interior angles is no more than 3.
Furthermore, since the polygon has three obtuse interior angles, the maximum number of sides is 6.
Therefore, the answer is (C).
[Solution 2] According to the problem, the $n$-sided polygon has 3 obtuse angles, so there are $n-3$ acute or right angles, thus
$$
(n-2) \cdot 180^{\circ} < (n-3) \cdot 90^{\circ} + 3 \cdot 180^{\circ} \text{, }
$$

Solving this, we get $n < 7$. Therefore, $n_{\max} = 6$.
Hence, the answer is (C).