Question

The angle θ lies in Quadrant I . cosθ=3/5

Answer 1

let's recall that in the I Quadrant, x,y or cosine,sine are both positive, thus

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-3^2}=b\implies \pm\sqrt{16}=b\implies \pm 4=b\implies \stackrel{I~Quadrant}{+4=b} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{4}}{\stackrel{adjacent}{3}}$