we know the angle is in the II quadrant, therefore the adjacent side or cosine is negative and the opposite side or sine is positive there.
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\
tan(\theta )=-\sqrt{3}\implies tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{3}}}{\stackrel{adjacent}{-1}}\qquad
\begin{array}{llll}
\textit{let's find the }\\
hypotenuse
\end{array}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf c=\sqrt{(-1)^2+(\sqrt{3})^2}\implies c=\sqrt{1+3}\implies c=\sqrt{4}\implies c=2\\\\
-------------------------------\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad sin(\theta)=\cfrac{\sqrt{3}}{2}[/tex]
