1. In $\triangle A B C$, it is known that: $a=2, b=2 \sqrt{2}$. Find the range of values for $\angle A$.
1. By the cosine rule: $\cos A=\frac{(2 \sqrt{2})^{2}+c^{2}-2^{2}}{2 \times 2 \sqrt{2} \cdot c}=\frac{1}{\sqrt{2} \cdot c}+\frac{c}{4 \sqrt{2}} \geqslant \frac{\sqrt{2}}{2}$, with equality if and only if $c=2$. Therefore, $\cos A \geqslant \frac{\sqrt{2}}{2}$, so $A \in\left(0, \frac{\pi}{4}\right]$.