## Precalculus (6th Edition) Blitzer

$\dfrac{\sqrt{3}}{3}$
RECALL: (1) $\sin{\theta} = \dfrac{\text{opposite side}}{\text{hypotenuse}}$ (2) $\tan{\theta} = \dfrac{\text{opposite side}}{\text{adjacent side}}$ (3) $\cos{\theta} = \dfrac{\text{adjacent side}}{\text{hypotenuse}}$ Use formula (1) above. Use the 30-degree angle of the triangle on the right to obtain: $\tan{30^o} = \dfrac{1}{\sqrt3}$ Rationalize the denominator by multiplying $\sqrt{3}$ to both the numerator and the denominator to obtain $\tan{30^o}=\dfrac{ 1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \\\tan{30^o}=\dfrac{\sqrt{3}}{3}.$