[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\
-------------------------------\\\\[/tex]
[tex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad \qquad sin(\theta_1)=\cfrac{11}{61}\cfrac{\leftarrow opposite=b}{\leftarrow hypotenuse=c}
\\\\\\
\textit{so, now we know who is the hypotenuse and opposite side}\\
\textit{let's find the adjacent side then}
\\\\\\
c^2=a^2+b^2\implies c^2-b^2=a^2\implies \pm\sqrt{c^2-b^2}=a
\\\\\\
\pm\sqrt{61^2-11^2}=a\implies \pm\sqrt{3600}=a\implies \pm 60=a[/tex]
but... which is it? the +/-? well, we know the angle is in the first quadrant, in the first quadrant cosine as well as sine, or "a" and "b" are both positive, so then a = 60
[tex]\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies cos(\theta_1)=\cfrac{60}{61}[/tex]
