[tex]\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)=\cfrac{\sqrt{7}}{2}\cfrac{\leftarrow hypotenuse=c}{\leftarrow opposite=b}\\\\
-----------------------------\\\\
\textit{so hmm using the pythagorean theorem, we can say that}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad
\begin{cases}
c=hypotenuse\\
b=opposite\\
a=adjacent
\end{cases}\\\\
\textit{we'll assume is the + of the root, since you're not}\\
\textit{given further info on it, that means the I quadrant}\\\\
[/tex]
[tex]\bf -----------------------------\\\\
cos(\theta)=\cfrac{a}{c}\qquad sec(\theta)=\cfrac{c}{a}\qquad tan(\theta)=\cfrac{b}{a}\qquad cot(\theta)\cfrac{a}{b}
[/tex]
so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant
keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant
but anyhow, use the I quadrant
