So, here we have the point:
![(-(11)/(61),(60)/(61))](https://img.qammunity.org/2023/formulas/mathematics/college/v5jhmemcr4txh8hvammw242jasyxr18sy5.png)
This is situated at the quadrant II.
Remember that sin(a) is a relation between the opposite side of the angle a and the hypotenuse of the triangle.
To find the hypotenuse, we apply the Pythagorean Theorem:
![\begin{gathered} h=\sqrt[]{((60)/(61))^2+((-11)/(61))^2} \\ h=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5j470qc32cevzzd9uhakrot8oq1naowl82.png)
Notice that as this point is in the unit circle, the hypotenuse is 1.
Now,
![\sin (\theta)=(60)/(61)](https://img.qammunity.org/2023/formulas/mathematics/college/itcnv6qwfz30b3ispnbjezitbys6ng5dfa.png)
And,
![sec(\theta)=(1)/(\cos (\theta))=(-61)/(11)](https://img.qammunity.org/2023/formulas/mathematics/college/qrv9vmvctnnpkjqyh5uekgk6e5oir3o9iu.png)
![\text{tan(}\theta)=((60)/(61))/((-11)/(61))=(-60)/(11)](https://img.qammunity.org/2023/formulas/mathematics/college/ugbetz2lbn0p6n5fz51klii23ge2gxynjf.png)