let's keep in mind that, in the II Quadrant, cosine is negative and sine, is positive.
cosine is adjacent/hypotenuse, however the hypotenuse is simply a radius unit, and thus is never negative, so in the -(2/3) the negative must be the numerator, -2.
[tex]\bf cos(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{hypotenuse}{3}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{3^2-(-2)^2}=b\implies \pm\sqrt{5}=b\implies \stackrel{\textit{II Quadrant}}{+\sqrt{5}=b}~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\sqrt{5}}}{\stackrel{adjacent}{-2}} \\\\\\ ~\hspace{34em}[/tex]
The value of tan θ will be -√5 / 2. Then the correct option is C.
The connection between the lengths and angles of a triangular shape is the subject of trigonometry.
The angle θ lies in Quadrant II .
cosθ = −2/3
The value of sin θ will be
[tex]\rm \sin \theta = \sqrt{1 - \cos^2 \theta}\\\sin \theta = \sqrt{1 - (-2/3)^2}\\\\[/tex]
sinθ = √5 / 3
Then the value of tan θ will be
tanθ = sinθ / cosθ
tanθ = (√5/3) / (-2/3)
tanθ = -√5 / 2
Then the correct option is C.
More about the trigonometry link is given below.
https://brainly.com/question/22698523
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