The quadrant for which cos Θ is negative is at Quadrant II and III.
The quadrant for which tan Θ is negative is at Quadrant II and IV.
The angle must be therefore found, in Quadrant II. Drawing a diagram of the angle, we have the following
Solve first for the missing side using the Pythagorean Theorem
[tex]\begin{gathered} a^2+b^2=c^2 \\ a^2+(-5)^2=(13)^2 \\ a^2+25=169 \\ a^2=169-25 \\ a^2=144 \\ \sqrt{a^2}=\sqrt{144} \\ a=12 \end{gathered}[/tex]
The missing side is 12, therefore,
[tex]\sin\theta=\frac{12}{13}[/tex]
