5. In a non-coplanar quadrilateral $A B C D$, the sum of the four interior angles is denoted as $\theta=\angle A B C+\angle B C D+\angle C D A+$ $\angle D A B$. Then $(\quad)$.
(A) $\theta>360^{\circ}$
(B) $\theta=360^{\circ}$
(C) $\theta<360^{\circ}$
(D) Cannot be determined, depends on the shape
5.C.

This means that the sum of the interior angles of a non-coplanar quadrilateral is less than $360^{\circ}$. The proof involves the properties of trihedral angles.

As shown in Figure 7, in quadrilateral $ABCD$, point $D$ is not in the plane $ABC$, and $AC$ is connected. By the property of trihedral angles that the sum of two face angles is greater than the third face angle, we have $\angle DAB < \angle BAC + \angle CAD$, $\angle BCD < \angle BCA + \angle DCA$. Adding these inequalities, and then adding $(\angle ABC + \angle CDA)$, we get
$$
\begin{array}{l}
\angle ABC + \angle BCD + \angle CDA + \angle DAB \\
< \angle ABC + (\angle BCA + \angle DCA) + \angle CDA + \\
(\angle BAC + \angle CAD) \\
= (\angle ABC + \angle BCA + \angle BAC) + \\
(\angle DCA + \angle CDA + \angle CAD) \\
= 180^{\circ} + 180^{\circ} = 360^{\circ}.
\end{array}
$$

Analysis: From this conclusion, we can deduce:
(1) A quadrilateral with four right angles must be a planar rectangle.
(2) A quadrilateral with three right angles is a rectangle. This conclusion holds in a plane but not in space. A counterexample is the quadrilateral $ABCD$ shown in Figure 8. This counterexample can negate (A) and (B), but not (D).