## Precalculus (6th Edition)

$\sqrt {3}x$
Lets assume that equation is $y=kx+b$ if the line passes through origin $b=0$ $k=\tan \alpha =\tan 60^{0}=\dfrac {\sin 60}{\cos 60}=\dfrac {\dfrac {\sqrt {3}}{2}}{\dfrac {1}{2}}=\sqrt {3}$ $\Rightarrow y=kx=\sqrt {3}x$