## Precalculus (6th Edition) Blitzer

$\sqrt3$
Convert the angle measure to degrees by multiplying $\dfrac{180^o}{\pi}$ to obtain: $\dfrac{\pi}{3} \cdot \dfrac{180^o}{\pi} = 60^o$ RECALL: (1) $\sin{\theta} = \dfrac{\text{opposite side}}{\text{hypotenuse}}$ (2) $\tan{\theta} = \dfrac{\text{opposite side}}{\text{adjacent side}}$ (3) $\cos{\theta} = \dfrac{\text{adjacent side}}{\text{hypotenuse}}$ (4) $\sec{\theta} = \dfrac{\text{hypotenuse}}{\text{adjacent}}$ (5) $\csc{\theta} = \dfrac{\text{hypotenuse}}{\text{opposite side}}$ Use formula (2) above. Use the 60-degree angle of the triangle on the right to obtain: $\tan{\frac{\pi}{3}} = \tan{60^o} = \dfrac{\sqrt3}{1}=\sqrt3$