3. As shown in the figure, it is known that APBCD is a figure symmetrical about the line $l$, and $\angle \mathrm{APD}=116^{\circ}$, $\angle \mathrm{DPC}=40^{\circ}$, $\mathrm{DC}>\mathrm{AB}$. Therefore, among all the triangles formed by the points $\mathrm{A} 、 \mathrm{P} 、 \mathrm{~B} 、 \mathrm{C}$ and D, there are __ obtuse triangles and __ acute triangles.
3. Solution: $C_{5}^{3}=\frac{5 \times 4 \times 3}{3 \times 2 \times 1}=10$, from the points $A, P, B, C, D$, 10 triangles can be formed. Among the interior angles of these 10 triangles,
$$
\angle \mathrm{APD}=\angle \mathrm{BPC}=116^{\circ}>90^{\circ}, \angle \mathrm{APC}=\angle \mathrm{BPD}=116^{\circ}+40=156^{\circ}>90^{\circ}
$$
Since $D C>A B$, $\angle A D C$ and $\angle B C D$ are acute angles, while $\angle B A D$ and $\angle A B C$ are obtuse angles,
$$
\angle \mathrm{APB}=360^{\circ}-116^{\circ} \times 2-40^{\circ}=88^{\circ}<90^{\circ},
$$

the rest are acute angles.
Therefore, there are 6 obtuse triangles and 4 acute triangles.