Look at the picture.
Definitions:
[tex]\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}[/tex]
We have:
[tex]\tan\theta=-\dfrac{2}{3}[/tex] and Quadrant II (x < 0, y > 0)
Therefore x=-3 and y = 2. Calculate r:
[tex]r=\sqrt{(-3)^2+2^2}=\sqrt{9+4}=\sqrt{13}[/tex]
Substitute to the formula of cosine:
[tex]\cos\theta=\dfrac{-3}{\sqrt{13}}=-\dfrac{3}{\sqrt{13}}\cdot\dfrac{\sqrt{13}}{\sqrt{13}}=-\dfrac{3\sqrt{13}}{13}[/tex]
Answer: a.
