Respuesta :
we know that the cos(θ) is -(3/5), however θ is in the II Quadrant, where the cosine is negative whilst the sine is positive, meaning the fraction is really (-3)/5, so
[tex]cos(\theta )=\cfrac{\stackrel{adjacent}{-3}}{\underset{hypotenuse}{5}}\qquad \qquad \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}[/tex]
[tex]\pm\sqrt{5^2-(-3)^2}=b\implies \pm\sqrt{25-9}=b\implies \pm 4=b\implies \stackrel{II~Quadrant}{+4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill csc(\theta )=\cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{4}}~\hfill[/tex]
