we have that
cos θ= 5/13 -----> given
θ is in quadrant 4 ----> that means
the value of the sine is negative and the value of cosine is positive
Remember that
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]substitute given value
[tex]\sin ^2\theta+(\frac{5}{13})^2=1[/tex][tex]\begin{gathered} \sin ^2\theta=1-\frac{25}{169} \\ \end{gathered}[/tex][tex]\sin ^2\theta=\frac{144}{169}[/tex][tex]\sin ^{}\theta=-\frac{12}{13}[/tex]