Example 9 In $\triangle A B C$, $A B=2, B C=1$, then the range of values for $\angle A$ is $\qquad$
(Suzhou City Junior High School Mathematics Competition)
Solution: Let $AC = x$. Then, by the cosine rule, we have
$$
BC^2 = AB^2 + AC^2 - AB \cdot AC \cdot \cos A,
$$

which is $1^2 = 2^2 + x^2 - 2 \times 2 \times x \cdot \cos A$, or equivalently, $x^2 - (4 \cos A) x + 3 = 0$.

Note that the problem now reduces to finding which values of $A$ make the above equation have real roots, i.e., the discriminant
$(4 \cos A)^2 - 4 \times 3 \geqslant 0$.
Solving this inequality, we get
$$
\cos A \leqslant -\frac{\sqrt{3}}{2} \text{ or } \cos A \geqslant \frac{\sqrt{3}}{2}.
$$

Since $AB = 2 > 1 = BC$, the angle $A$ opposite to $BC$ must be acute. Therefore, we have
$$
0^\circ < \dot{A} \leqslant 30^\circ.
$$
(3) Solving a geometric discussion problem