Respuesta :

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]

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