$7 \cdot 48$ In an acute triangle $\triangle ABC$, the three sides are $a, b, c$, and the distances from its circumcenter to the three sides are $m, n, p$. Then, $m: n: p$ equals
(A) $\frac{1}{a}: \frac{1}{b}: \frac{1}{c}$.
(B) $a: b: c$.
(C) $\cos A: \cos B: \cos C$.
(D) $\sin A: \sin B: \sin C$.
(China Junior High School Mathematics League, 1993)
[Solution 1] Connect $O A, O B, O C$, then $O A=O B=O C=R$ (circumradius). Clearly, by the Pythagorean theorem we have
$$
\begin{array}{l}
m^{2}=R^{2}-\left(\frac{a}{2}\right)^{2}=R^{2}-\frac{a^{2}}{4}, \\
n^{2}=R^{2}-\frac{b^{2}}{4} \\
p^{2}=R^{2}-\frac{c^{2}}{4}
\end{array}
$$

Thus, $m: n: p=\sqrt{R^{2}-\frac{a^{2}}{4}}: \sqrt{R^{2}-\frac{b^{2}}{4}}: \sqrt{R^{2}-\frac{c^{2}}{4}}$
(divide by $R$ )
$$
\begin{array}{l}
=\sqrt{1-\left(\frac{a}{2 R}\right)^{2}}: \sqrt{1-\left(\frac{b}{2 R}\right)^{2}}: \sqrt{1-\left(\frac{c}{2 R}\right)^{2}} \\
=\sqrt{1-\sin ^{2} A}: \sqrt{1-\sin ^{2} B}: \sqrt{1-\sin ^{2} C} \\
=\cos A: \cos B: \cos C .
\end{array}
$$

Here, note that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$, and $\triangle A B C$ is an acute triangle.
Therefore, the answer is $(C)$.
[Solution 2] Connect $O A, O B, O C$, clearly
$$
O A=O B=O C=R \text { (circumradius). }
$$

Also, $\angle C O B=2 \angle A$, and $O M$ is perpendicular and bisects $B C$, so $O M$ bisects $\angle C O B$.
In the right triangle $\triangle C O M$, $\angle C O M=\frac{1}{2} \angle C O B=$ $\angle A$,
then $m=R \cos A$.
Similarly, $n=R \cos B, p=R \cos C$.
$$
\begin{aligned}
\therefore m: n: p & =R \cos A: R \cos B: R \cos C \\
& =\cos A: \cos B: \cos C .
\end{aligned}
$$

Therefore, the answer is $(C)$.