Respuesta :
keep in mind that, in the IV quadrant, sine or the y-coordinate is negative, and the cosine or x-coordinate is positive, whilst the hypotenuse or radius, is just a distance unit and is never negative.
now, we know the angle is in the IV quadrant, therefore the opposite side or "y" is negative, thus
[tex]\bf sin(\theta )=-\cfrac{9}{41}\implies sin(\theta )=\cfrac{\stackrel{opposite}{-9}}{\stackrel{hypotenuse}{41}} \\\\\\ \textit{so, now let's find the \underline{adjacent side} then} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf \pm\sqrt{41^2-(-9)^2}=a\implies \pm\sqrt{1600}=a\implies \pm 40=a \\\\\\ \stackrel{\textit{in the IV quadrant}}{+40=a}\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad tan(\theta)=\cfrac{-9}{40}[/tex]
now, we know the angle is in the IV quadrant, therefore the opposite side or "y" is negative, thus
[tex]\bf sin(\theta )=-\cfrac{9}{41}\implies sin(\theta )=\cfrac{\stackrel{opposite}{-9}}{\stackrel{hypotenuse}{41}} \\\\\\ \textit{so, now let's find the \underline{adjacent side} then} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}[/tex]
[tex]\bf \pm\sqrt{41^2-(-9)^2}=a\implies \pm\sqrt{1600}=a\implies \pm 40=a \\\\\\ \stackrel{\textit{in the IV quadrant}}{+40=a}\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad tan(\theta)=\cfrac{-9}{40}[/tex]
