Let $O$ be a unit circle which means its radius $OA=OB=1$. If $\angle{DOA}=x$ (in radian), then arc $\widehat{AB}=x$ by the definition of radian measurement. Meanwhile, based on the definitions of trigonometric functions, we find segment $BC=\sin x$, and $DA = \tan x$. It is intuitive to see that the areas satisfy $$S_{\triangle{OAB}} < S_{OAB} < S_{\triangle{OAD}} \Leftrightarrow \frac{1}{2}\cdot 1 \cdot \sin x < \pi \cdot 1^2 \cdot\frac{x}{2\pi} < \frac{1}{2} \cdot 1 \cdot \tan{x} \implies \sin x < x < \tan x$$