[tex]\frac{5\sqrt[]{39}}{39}[/tex]
1) We need to make use of one Pythagorean Trig Identity so that we can find the cotangent of A.
[tex]\begin{gathered} \sin ^2(A)+\cos ^2(A)=1 \\ \sin ^2(A)+(\frac{5}{8})^2=1 \\ \sin ^2(A)=1-\frac{25}{64} \\ \sin (A)=\sqrt[]{\frac{64}{64}-\frac{25}{64}} \\ \sin (A)=\sqrt[]{\frac{39}{64}}=\frac{\sqrt[]{39}}{8} \end{gathered}[/tex]
2) Now, we can have the cotangent since we know the sine, the cosine:
[tex]\cot (A)=\frac{\cos(A)}{\sin(A)}=\frac{\frac{5}{8}}{\frac{\sqrt[]{39}}{8}}=\frac{5}{8}\times\frac{8}{\sqrt[]{39}}=\frac{5}{\sqrt[]{39}}\cdot\frac{\sqrt[]{39}}{\sqrt[]{39}}=\frac{5\sqrt[]{39}}{39}[/tex]
