[tex]\begin{gathered} sin\text{ }\theta\text{ = }\frac{-5}{13} \\ sin\theta\text{ =}\frac{Opposite}{Hypotenuse} \\ Hypotenuse^2\text{ = Opposite}^2\text{ + Adjacent}^2 \\ Adjacent\text{ = }\sqrt[]{13^2-(-5)^2} \\ =\text{ 12} \\ If\text{ tan}\theta\text{ }>0\text{ and sin}\theta<0 \\ This\text{ means that the angle is in quadrant three where cos is also negative} \\ cos\theta\text{ = }\frac{adjacent\text{ }}{hypotenuse} \\ \text{ = }\frac{-12}{13}\text{ } \end{gathered}[/tex]
Correct option number 3.
