we know that sin(θ) is 4/5, and we also know that θ is in the I Quadrant, where sine and cosine are both positive, so
[tex]sin(\theta )=\cfrac{\stackrel{opposite}{4}}{\underset{hypotenuse}{5}}\qquad \qquad \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-4^2}=a\implies \pm 3=a\implies \stackrel{I~Quadrant}{+3=a}~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}[/tex]
