Side-Angle Inequalities in a Triangle

It is intuitive that larger angles mean greater sides, but let us try to prove this.

The Law of Sines tells us $a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.

We have to split into two cases: ABC is obtuse and ABC is acute.

If ABC is acute, then the result is obvious, because since all angles are less than 90, we have $\sin x > \sin y \iff x > y$

If ABC is obtuse, then the result is still obvious. Suppose C is obtuse. Then $\sin C = \sin (180 - C)$

Thus, since $180 - C$ is greater than A and B, the same argument holds.

First I want us to imagine to imagine a triangle ABC. Now remove side BC. Now we have two segments AB and AC and an angle BAC. So say we that we decreased the measure of angle BAC. Now obviously the side BC has shorter length than before. Now repeat the same process, except this time we make angle BAC larger. This results in BC having larger length. This demonstrates in a triangle, longer sides are opposite larger angles, and vice versa.