Respuesta :
hmmm if you don't have a Unit Circle, this is a good time to get one, many you can find online. Anyhow, check your unit circle for cos(30°) and sin(30°).
[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad tan(30^o)=\cfrac{sin(30^o)}{cos(30^o)} \\\\\\ tan(30^o)=\cfrac{\quad \frac{1}{2}\quad }{\frac{\sqrt{3}}{2}}\implies tan(30^o)=\cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies tan(30^o)=\cfrac{1}{\sqrt{3}} \\\\\\ \stackrel{\textit{and now if we rationalize the denominator}}{\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{(\sqrt{3})^2}\implies \cfrac{\sqrt{3}}{3}}[/tex]
[tex]\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad tan(30^o)=\cfrac{sin(30^o)}{cos(30^o)} \\\\\\ tan(30^o)=\cfrac{\quad \frac{1}{2}\quad }{\frac{\sqrt{3}}{2}}\implies tan(30^o)=\cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies tan(30^o)=\cfrac{1}{\sqrt{3}} \\\\\\ \stackrel{\textit{and now if we rationalize the denominator}}{\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{(\sqrt{3})^2}\implies \cfrac{\sqrt{3}}{3}}[/tex]