5. If $a, b, c$ are real numbers, then for any real number $x$, the necessary and sufficient condition for the inequality $a \sin x + b \cos x + c > 0$ to always hold is
(A) $a, b$ are both 0, and $c > 0$
(B) $\sqrt{a^{2}+b^{2}}=c$
(C) $\sqrt{a^{2}+b^{2}}<c$
$a \sin x+b \cos x+c$
$$
=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)+c.
$$

Here, $\operatorname{tg} \varphi=\frac{b}{a}$.
However, when making the equivalent transformation
$$
\begin{array}{l}
a \sin x+b \cos x+c>0 \\
\Leftrightarrow \sin (x+\varphi)>-\frac{c}{\sqrt{a^{2}+b^{2}}}
\end{array}
$$

some students feel confused and make the erroneous transformation
$$
\begin{array}{l}
\sin (x+\varphi)>-\frac{c}{\sqrt{a^{2}+b^{2}}}= \\
\Leftrightarrow-\frac{c}{\sqrt{a^{2}+c^{2}}}>-1.
\end{array}
$$

Thus, they incorrectly choose (D).
Below, we provide the geometric interpretation of the original inequality. For this, let $\theta$ be any angle. The inequality $a \sin \theta+b \cos \theta+c>0$ actually means that any point $(\cos \theta, \sin \theta)$ on the unit circle lies above the line
$$
\text { l: } b x+a y+c=0
$$
with a distance of 1, i.e.,
$$
\left|\frac{c}{\sqrt{a^{2}+b^{2}}}\right|>1.
$$

The center of the circle also satisfies the inequality $b x+a y+c>0$, i.e., $c>0$.
Therefore, $c>\sqrt{a^{2}+b^{2}}$.
Based on this, the correct choice is (C).
With the geometric interpretation, the erroneous equivalent relation (3) would not occur.
Of course, if the basic concepts are clear and the skills are proficient, no mistakes would be made.
In fact, in (2), if we take $x=-\frac{\pi}{2}-\psi$, we get
$-1>-\frac{c}{\sqrt{a^{2}+b^{2}}}$,
i.e., $c>\sqrt{a^{2}+b^{2}}$.
Conversely, from this inequality and $\sin (x+\varphi)>-1$, we immediately obtain (2). Therefore, the correct choice is (C).