An angle θ terminates in quadrant II. If tan θ = - squareroot3, determine the value of sin θ.

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]