we know the angle is between 90° and 180°, 90° < θ < 180°, that means the angle is in the II Quadrant, where the sine is positive and the cosine is negative, with that info, let's find check a few things
[tex]sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\underset{hypotenuse}{5}}\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 - 3^2}=a\implies \pm\sqrt{16}=a\implies \pm 4 = a\implies \stackrel{II~Quadrant}{-4=a} \\\\[-0.35em] ~\dotfill[/tex]
[tex]cos(\theta )=\cfrac{\stackrel{adjacent}{-4}}{\underset{hypotenuse}{5}} \\\\\\ tan\left(\cfrac{\theta}{2}\right)\implies \cfrac{sin(\theta)}{1+cos(\theta)}\implies \cfrac{ ~~ \frac{3}{5} ~~ }{1-\frac{4}{5}}\implies \cfrac{ ~~ \frac{3}{5} ~~ }{\frac{1}{5}}\implies \cfrac{3}{5}\cdot \cfrac{5}{1}\implies 3[/tex]
