Track / Stage 4 / 237 of 340 #497 of 1964

Problem 497

AMC 12 late, AIME early

Geometry Difficulty 4.9 Find the answer

1. In $\triangle A B C$ , it is known that: $a=2, b=2 \sqrt{2}$ . Find the range of values for $\angle A$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Official solution

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 $A \in\left(0, \frac{\pi}{4}\right]$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.