well, we know the angle is acute, that means is less than 90°, meaning is in the first quadrant, that means the "x" or adjacent side as well as the "y" or opposite side, are both positive.
[tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\
cos(a)=\cfrac{\sqrt{7}}{7}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\quad \textit{now, let's find the \underline{opposite}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf \pm\sqrt{7^2-(\sqrt{7})^2}=b\implies \pm\sqrt{49-7}=b\implies \pm\sqrt{42}=b
\\\\\\
\textit{now, which is it, the + or -? well, we're in the 1st quadrant, is }\sqrt{42}
[/tex]
[tex]\bf
\textit{now that you know all three sides, just plug them in}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\\\\\\
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\qquad
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\quad
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}[/tex]
